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Similar Figures and Trigonometry SIMILAR FIGURES Key Ideas
Definition of Similarity and Proportional Parts Two figures that have the shape but possibly different sizes are called similar. You can think of similar figures as a photocopy enlargement or reduction of one another. Any two similar figures must have certain properties.
Properties of Similar Figures All corresponding lengths are proportional. All corresponding angles are congruent. The constant of proportionality is called the scale factor. Figure 8.1 shows two similar quadrilaterals, ABCD and WXYZ. We can write a similarity statement using the similarity symbol ~.
When pairing the corresponding vertices, the similarity statement is ABCD ~ WXYZ.
The four pairs of congruent corresponding angles are as follows: Similar quadrilaterals ABCD and WXYZ
The four corresponding pairs of proportional sides are the following:
Each of the ratios is equivalent to , so the scale factor is . Math Fact It’s not just corresponding sides that are proportional in similar figures. Any two segments whose endpoints are corresponding points will be proportional. For example in Figure above, diagonal lengths AC and WY are also in a 1 : 2 ratio. In similar triangles, all corresponding lengths must be proportional. Example 1 ΔPST has side lengths PS = 3, ST = 4, and TP = 5. ΔBLM has side lengths BL = 9, LM = 12, and MB = 15. If the two triangles are similar, what is the scale factor? Write a similarity statement for the two triangles. Solution: Find the scale factor by writing the ratios of corresponding sides and pairing them up so all three ratios are equal. To be sure you are forming ratios with the correct parts, pair up the smallest side from each figure, then the next biggest, and so on.
The three ratios are the following: Each ratio simplifies to , which is the scale factor. The similarity statement is ΔPST ~ ΔBML. The two largest sides are TP and BL. The two medium sides are ST and LM. The two smallest sides are PS and MB. These pairs match up to the corresponding sides as specified in the similarity statement. Example 2 Given similar triangles ΔABC ~ ΔJKL, m∠A = 102°, and m∠K = 43°. Find m∠B and m∠C. Solution: The strategy here is to combine two theorems. From the similarity relationship, we know corresponding angles are congruent. From the triangle angle sum theorem: Finding a Missing Side If you know three out of the four parts of a proportion made from corresponding parts, you can solve for the fourth. Use the fact that the cross products of a proportion are equal to write an equivalent equation. Example 1 Given ABCDE ~ VWXYZ, with AB = 18, BC = 30, and WX = 20. Find the length of WV. Solution: From the similarity statement, and are corresponding parts, as are and . The equation that results from solving the proportion may be quadratic. In that case, use any technique you are most comfortable with to solve the quadratic, such as factoring, the quadratic formula, completing the square, or trial and error. Also look out for given information that may give a clue to the scale factor, such as midpoints. Example 2 Rectangle TRIP is similar to rectangle BSAP. A is the midpoint of , RI = x2, SA = x + 4, and TR = 10x. Find BS. Solution: Midpoint A divides in half, so .
This gives us the scale factor between the two rectangles. Scaled Drawing Similar figures, scale factors, and dilations are frequently used by architects, engineers, graphic designers, and many other professions. An engineer or an architect may create a design on paper of a building, car, or any other object. The drawing is called a scaled drawing because it is similar to the actual object it represents. All lengths are reduced or enlarged by the same scale factor. The scaled drawings let you see what the final object will look like, give guidance to someone who will build the object, and help designers determine how multiple parts fit together. When working with a scaled drawing, apply the similar figure relationships between corresponding lengths, angles, area, and volume. Example An interior designer is sketching a new layout for a living room that calls for installing carpet in the room. The room is rectangular in shape, and her sketch measures 12 inches by 9 inches. If the scale factor on the drawing is 1:20 and carpet costs $3.99 per square foot, how much should she budget for the carpet?
Solution: First find the actual dimension of the room. Now convert the area to square feet so the cost can be calculated. Note that the scale factor is squared when converting in.2 to ft2. The total cost will be the area multiplied by the cost per square foot. The designer should budget $1,197 for the carpet. Scaling Perimeter, Area, and Volume Looking back to the carpet problem in the previous section, there is a relationship between the area of the drawing, the actual area of the room, and the scale factor. The area of the room in the drawing is 12 in. · 9 in. = 108 in.2
The ratio of the two areas is
This ratio is exactly equal to the scale factor squared:
Whenever a figure undergoes a similarity transformation, the areas of the preimage and image are in a ratio equal to the scale factor squared.
This is because area will always be equal to the product of two lengths and each length is multiplied by the scale factor. Volume will always be the product of three lengths. So the volume of two similar solids is proportional to the scale factor cubed.
Example 1 ΔSKY ~ ΔBLU, SK = 12, BL = 3, and the area of ΔBLU is 5 cm2. What is the area of ΔSKY? Solution: The ratio of the areas is equal to the square of the ratio of corresponding sides. Use the corresponding lengths SK and BL. Example 2 The volume of a rectangular prism is 10 in.3 If the prism undergoes a dilation with a scale factor of 4, what is the volume of the dilated prism? Solution: Every dimension will be multiplied by 4. So the ratio of any pair of corresponding lengths will be 4. Volume is proportional to scale factor cubed. Example 3 Kate wants to buy Styrofoam spheres for a craft project. Last week, she bought 3 in. diameter spheres for $0.75 each. This week, she wants to buy 8 in. spheres. Estimate the cost for each 8 in. sphere, rounded to the nearest dollar. Justify your reasoning. Solution: The scale factor is the ratio of the diameters, or Volume is scaled by a factor of .
A reasonable estimate for the cost is to scale the cost in the same way as the volume since the amount of material in the sphere is proportional to the volume. Example 4 A painter uses 8 gallons of paint to paint a house. At his next job, he estimates the length and width of the house are twice that of the first house he painted. Approximately how much paint should he expect to use? Solution: The amount of paint will be proportional to the surface area of the house, and surface area is proportional to the scale factor of the length squared. In this case, the scale factor is 2.
PROVING TRIANGLES SIMILAR AND SIMILARITY TRANSFORMATIONS Key Ideas Two triangles can be proven similar using the following postulates: AA (angle-angle), SAS (side-angle-side), and SSS (side-side-side) Proving Triangles Similar by the Angle-Angle Criterion The Angle-Angle (AA) Similarity Criterion If two pairs of corresponding angles are congruent in a pair of triangles, then the triangles are similar. The AA criterion can be justified by showing a similarity transformation exists that maps one triangle to the other. From a transformation point of view, two figures are similar if a sequence of similarity transformations maps one figure to another. Figure 8.2a shows ΔABC and ΔDEF with two pairs of corresponding congruent angles.
A translation by vector will map point D to A, shown in Figure 8.2b. A rotation about point A by ∠CAF will map F′ onto side and E′ on side , as shown in Figure c. Finally, a dilation about point A by scale factor will complete the mapping. Since a dilation was used, the triangles are not necessarily congruent, but they are similar. ΔABC ~ ΔDEF by the AA criterion Math Fact Having all three pairs of corresponding angles congruent in a pair of triangles is sufficient to prove them similar. (In fact, we just saw that having even two pairs is enough.) However, having all pairs of corresponding angles congruent in two polygons with more than three sides does not prove they are similar.
For example, a square and a rectangle have all right angles, but they are not necessarily similar. As with other proofs, vertical angles and parallel lines are often used when proving two triangles are similar. Also take care when writing a similarity statement. The corresponding parts must be listed in the same order. Example 1 Given: and intersect at E, , and Prove: ΔRDE ~ ΔBSE Also state a sequence of rigid motions that will map ΔRDE onto ΔBSE.
Solution: Use the pair of vertical angles and pair of right angles to show the triangles are similar by AA.
A rotation of 180º about point E followed by a dilation centered at E with scale factor will map ΔRDE onto ΔBSE. Example 2 Given: and intersects at Z Prove: ΔZXY ~ ΔZQP Solution: Use the congruent alternate interior angles to show two angles are congruent.
Example 3 Given: , ∠A ≅ ∠S, AB = 8, and RS = 14
Prove: Solution: ∠A ≅ ∠S is given. Vertical angles ∠ACB and ∠RCS are congruent. Therefore, ΔACB ~ ΔSCR by the AA criterion. Corresponding parts of similar triangles are proportional, so .
Substituting the given lengths for RS and AB yields , which simplifies to . Proving Triangles Similar by the Side-Angle-Side Criterion If two pairs of corresponding sides are proportional in a pair of triangles and the included angles are congruent, then the triangles are similar. Example In ΔCAR, CA = 9 and AR = 12. ΔTOY has side lengths TO = 30 and TY = 40. m∠A = m∠O = 18°. Explain why ΔCAR and ΔTOY are similar. Solution: The two pairs of given sides are proportional. The included angles in each triangle are ∠A and ∠T. These both have the same measure, so they are congruent. The triangles are similar by the SAS criterion. Proving Triangles Similar by the Side-Side-Side Criterion If three pairs of corresponding sides are proportional in a pair of triangles, then the triangles are similar. Figure 8.3 shows ΔABC ≅ ΔDEF by the SSS criterion. ΔABC ~ ΔDEF by the SSS criterion Example In the accompanying figure, HI = 12, IP = 12, HP = 8, SH = 8, and . Prove ΔHOP ~ ΔSHP.
Solution: By matching up corresponding sides in order of length, we have: The first two ratios simplify to . The third ratio also simplifies to , although that one is not quite as obvious. One way to confirm that two ratios are equal is to show that the cross products are equal.
Since all three pairs of corresponding sides are proportional, the triangles are similar by SSS. Similarity Transformations As discussed previously, the dilation is a similarity transformation. If given a pair of similar figures, one can always be mapped to another by a similarity transformation that will include a dilation. The concept is very much like demonstrating congruence by transformations, except that there will be a dilation to match the sizes of the figures. It is easiest to first map the two vertices that will be the center of the dilation onto each other. That way when the dilation is applied, those two vertices will still be coincident.
In the following examples, we will combine multiple aspects of similarity: proving figures similar, finding a missing side using a similarity relationship, and finding a transformation that maps one similar figure to another. Example 1 Given: and intersecting at S, , IP = 4, and FL = 6 Prove the two triangles formed are similar. Find the ratio .
State a specific similarity transformation that will map the smaller triangle onto the larger. Solution: Given , we know ∠F ≅ ∠P and ∠L ≅ ∠I because they are the alternate interior angles formed by parallel lines and a transversal. Therefore ΔSIP ~ ΔSLF by AA. (Note the correct order of vertices in the similarity statement.)
Since corresponding parts are in the same ratio: A rotation of 180° about point S will align with and align with . A dilation with center at S and a scale factor of will then map ΔSIP to ΔSLF. Note that we have to use the reciprocal of the ratio in part (b) because we want to enlarge the smaller triangle, not reduce it. In transformation notation, the required notation is Example 2 ABCD is a rectangle, and . AF = 15, FB = 3, BG = 4, and GC = 6. Prove ΔADE ~ ΔGBF. Find DE. What transformation maps ΔGBF to ΔADE? Solution: Opposite angles D and B are the first pair of congruent angles. To show the second pair is congruent, we need to use the two pairs of parallel lines, and .
and are congruent corresponding sides, as are and . Opposite sides of a rectangle are congruent, so AD = BC = 10.
We can form a proportion: Three steps are needed to map ΔGBF to ΔADE. First translate ΔGBF by vector . This maps B to D. Next rotate 180° about point D to align with and align with .
Finally a dilation is needed. The scale factor is , or , and the center is point D. The complete transformation can be expressed symbolically as .
SIMILAR TRIANGLE RELATIONSHIPS Segment Parallel to a Side of a Triangle Segment Parallel to a Side Theorem A segment parallel to a side of a triangle forms a triangle similar to the original triangle. If a segment intersects two sides of a triangle such that a triangle similar to the original is formed, the segment is parallel to the third side of the original triangle. Figure shows ΔABC with parallel to . ΔPQC is similar to ΔABC by the AA postulate. ∠BAC and ∠QPC are congruent corresponding angles formed by the parallel lines. The same is true for ∠ABC and ∠PQC. Two pairs of corresponding angles are congruent, so the triangles are similar by AA. Segment side
The triangles can be sketched separately to help identify corresponding parts. Sometimes segment addition is needed to find a length needed to complete a proportion involving corresponding sides. Example In ΔWXY, is parallel to , RY = 8, XR = 12, and XW = 25. Find RS. Solution: Start by sketching the two triangles separately and using segment addition to find XY. The similarity relationship is ΔWXY ~ ΔSRY. The proportion of corresponding sides is as follows: Side Splitter Theorem A segment parallel to a side in a triangle divides the two sides it intersects proportionally. Figure shows parallel to and dividing sides and proportionally.
Specifically: Parallel segment divides and proportionally
The side splitter theorem is a shortcut that you can use to find a missing side when there is a line or segment parallel to a side of a triangle. You can prove the side splitter theorem by using the relationship and then applying the partition postulate. The important thing to remember is that it is the two sides intersected by the parallel segment that are divided proportionally. The ratio cannot be combined with the ratio . Example 1 In ΔHRS, , OH = (x + 1), OR = 2, EH = (x + 3), and ES = 2.5. Find the length of HR. Solution: Side divides the two sides proportionally according to the side splitter theorem. The side splitter theorem also applies if multiple parallel lines are intersected by two transversals. Figure 8.6 shows 4 parallel lines intersected by two transversals. Any corresponding pairs of segments will be proportional:
Side splitter theorem applied to multiple parallel lines intersected by two transversals Example 2 In Figure 8.6, a = 12, c = (x + 4), d = 18, and f = (x + 7). Find c. Solution: Use the side splitter theorem: Centroid Recall that the median of a triangle is a segment from a vertex of the triangle to the midpoint of the opposite side. Also that the medians in a triangle are all concurrent at the centroid. The centroid has the special property that it divides each median in a 1 : 2 ratio. Centroid Theorem The centroid of a triangle divides each median in a 1 : 2 ratio, with the longer segment having a vertex as one of its endpoints. If you are given the length of either the shorter or longer piece of a median, simply multiply or divide by 2 to find the other piece. If you are given the length of the entire median, the 1 : 2 ratio allows you to assign x to the shorter piece and 2x to the longer piece. Applying the whole equals the sum of the parts results in: Example 1 In ΔPQR, medians , and intersect at G. PM = 12, QG = 9, and OG = 2. Find PG Find QN Find RO Solution: Each median is divided in a 1 : 2 ratio. Let PG = 2x and GM = x: QG is the longer part of median : OG is the shorter part of median : You may not be told explicitly that a segment is a median. However, if its endpoints are a vertex and the midpoint of the opposite side, then it must be a median. If you are given expressions for each part of the median in terms of a variable, set up a proportion using the 1 : 2 ratio and then solve the proportion for the variable. Example 2 In the accompanying figure of ΔABC, D is the midpoint of and E is the midpoint of . If GC = (6x − 12) and GD = (2x + 4), find the length of .
Solution: and are medians, so G is a centroid and divides in a 1 : 2 ratio: Now substitute x = 10 and solve for GC and GD:
Add the two parts to find CD:
Midsegments Definition A midsegment is a segment joining the midpoints of two sides of a triangle. Midsegment Theorem A midsegment of a triangle is parallel to the opposite side and its length is equal to the length of the opposite side. The midsegment relationships are easily proven using the theorem that states that corresponding parts of similar triangles are proportional.
Figure shows midsegment in ΔABC. ΔABC can be proven similar to ΔADE using the SAS similarity postulate. Since D and E are midpoints, and , resulting in two pairs of proportional sides.
The included angle, ∠A, is a shared angle. Therefore ΔABC ~ ΔADE by SAS. If given expressions for a midsegment and the opposite side, use the following relationship:
Midsegment in ΔABC Example 1 Points M and N are the midpoints of and in ΔDRQ. If MN = (x + 5) and DR = (6x − 6), find the length of . Solution: First sketch the figure.
We see midsegment is opposite side :
Example 2 In ΔBIG, the midpoint of side is P and the midpoint of side is Q. If m∠QPI = 38° and m∠BGI = 41°, what is the measure of ∠I ? Solution: It is best to sketch the figure first.
is a midsegment and is parallel to . ∠B and ∠QPI are congruent corresponding angles, so m∠B = 38°. We can now use the triangle angle sum theorem in ΔBIG to find m∠I.
Similarity in Right Triangles Altitude to a Hypotenuse Theorem An altitude to the hypotenuse of a right triangle will divide it into two similar triangles, each of which is also similar to the original right triangle. In Figure of right ΔABC, altitude is drawn to hypotenuse . Three similar triangles are formed—ΔCBD, ΔACD, and the original ΔABC. Altitude drawn to hypotenuse in right ΔABC
When given any three parts, or sometimes two parts, any of the other parts can be found by combining the fact that corresponding parts are proportional and by using the Pythagorean theorem. Two useful strategies for dealing with these overlapping similar triangles is to sketch the triangles separately or to make a table with sides classified as “leg 1,” “leg 2,” and “hypotenuse.” The altitude will be leg 1 of one of the interior triangles and will be leg 2 of the other. Example 1 Altitude is drawn to hypotenuse in right triangle . If LU = 15 and LE = 9, what is the length of BL? Solution: Start by filling out a table of each of the sides of the three similar triangles. Let LE be leg 1 of ΔLUE and leg 2 of ΔBLE. (It won’t matter if you switch them.)
To make a proportion, we need two matching pairs of corresponding parts from two triangles. We don’t have the necessary two pairs. However, the table shows that if we can find leg 2 of ΔLUE (side EU), we can form a proportion to find BL.
We can find EU using the Pythagorean theorem:
We can now apply the proportional sides relationship with the following sides:
Math Fact When three numbers satisfy the proportion b is said to be the “mean proportional” between a and c. In geometry, the mean proportional shows up in similar right triangles. The altitude to the hypotenuse in a right triangle is the mean proportional to the two parts of the hypotenuse. Example 2 is the altitude to hypotenuse of ΔPNM. If PO = 12 and MO = 16, what is the length of ? Express your answer in simplest radical form.
Solution: ON is the mean proportional between OP and MO. The proportion is You would arrive at the same proportion using the table method; the proportion is equivalent to . Proving the Pythagorean Theorem One of the required proofs in the Common Core Curriculum is proving the Pythagorean theorem. There are many ways to prove this theorem, but the method you need to be able to use is one using similar triangles. We just saw the theorem that an altitude to the hypotenuse of a right triangle divides the triangle into two triangles similar to the original. This theorem can be used to prove the familiar a2 + b2 = c2. The strategy is to write two proportions relating corresponding sides. The first proportion will use the original triangle and one interior triangle (ΔABC and ΔBDC in the figure below). The second proportion uses the original triangle and the other interior triangle (ΔABC and ΔCDA). These can be rearranged algebraically and added to give the desired result.
Given: right triangle ΔABC with a right angle at C, altitude drawn to hypotenuse Prove: AC 2 + BC 2 = AB2
RIGHT TRIANGLE TRIGONOMETRY Definition of Trigonometric Ratios From the properties of similar figures, we know that the ratio of any two sides in one triangle will be equal to the ratio of two corresponding sides in a similar triangle. Right triangles occur so frequently in practical applications that mathematicians have given names to the parts of a right triangle and to the ratios formed by the sides of a right triangle. We refer to the sides of a right triangle relative to a particular acute angle using the names adjacent, opposite, and hypotenuse. Figure 8.9 shows the position of the adjacent, opposite, and hypotenuse for the two acute angles A and B in ΔABC. (a) Adjacent, opposite, and hypotenuse relative to angle A (b) Adjacent, opposite, and hypotenuse relative to angle B
Hypotenuse—the side across from the right angle Opposite—the side across from the specified acute angle Adjacent—the side included between the right angle and the specified acute angle
Right triangle trigonometry in this course will focus the following special ratios of sides in right triangles.
When working with trigonometric ratios, keep the following facts in mind: Sine, cosine, and tangent are functions that depend on the input angle. Sine, cosine, and tangent will have the same value for a given angle, regardless of the side lengths of the triangle. The sine and cosine ratios will always be greater than 0 and less than 1. The tangent ratio can be greater that 1. Math Fact The power of trigonometry is that these ratios will always be the same for a particular angle in a right triangle, regardless of how big or small the triangle is. This occurs because two right triangles with one pair of congruent acute angles must be similar by the AA postulate. By just knowing the angle, we know the ratio of two sides. Example 1 Find sin(C), cos(C), and tan(C) in the accompanying triangle.
Solution: Note that the radical expressions for sin(C) and cos(C) were rationalized by multiplying the numerator and denominator by in the last step. The Pythagorean theorem can be used to find a missing side that may be needed to form a trigonometric ratio. Example 2 In ΔABC, m∠B = 90°, BC = 6, and AC = 10. Find the value of cos(A). Solution: Sketching the figure will help identify the proper parts. We need the adjacent side, AB, to form the cosine ratio. By applying the Pythagorean theorem, we get: When you are given a trigonometric ratio but no side lengths, you can assume any side lengths consistent with the given ratio. Any similar right triangle with sides in the same proportion will have the same angles and trigonometric ratios. Example 3 In ΔFGH, m∠H = 90° and . What is tan(F)? Solution: Start by sketching ΔFGH. Let GH = 2 and FG = 3. The resulting triangle will be similar to any other right triangle with the same angle measures. Example 4 ΔPQR has a right angle at Q, sin(P) = a, and cos(P) = b. Find the values of sin(R) and cos(R). Solution: Sketch and label the figure. Then choose values for the opposite and hypotenuse that are consistent with sin(P) = a. Letting QR = a and PR = 1 is an obvious (but not the only) choice. Since cos(P) must equal b and the hypotenuse equals 1, then PQ must equal b. Calculator Tip sin, cos, and tan keys. Most calculators can be set to different units of angle measure, such as the degree, radian, or grad. It is important to know which unit the angle is given in. If the angle is followed by the degree symbol °, you are working in degrees, so the calculator should be in degree mode. For the TI-84 family of calculators, for example, you can switch between degrees and radians by pressing mode and using the arrow keys to select degree or radian.
Remember—Some graphing calculators are automatically set to radian mode whenever the memory is reset! You may need to set it back to degree mode. Finally, when doing trigonometric calculations using the calculator, do not round until you have reached the final answer. This is good practice for any problem involving numerical calculations. Example 5 In ΔABC, ∠C is a right angle and m∠A = 41°. Find the ratios , and . Round to the nearest thousandth. Solution: A sketch of the triangle shows that is the opposite of ∠A, is the adjacent of ∠A, and is the hypotenuse.
Inverse Trigonometric Functions The inverse of a function “undoes” the original function.
For example, and g(x) = x2 are inverse functions. If x = 5, then .
The trigonometric functions sin(x), cos(x), and tan(x) also have inverses, called the arcsin(x), arccos(x), and arctan(x).
They are often abbreviated sin−1(x), cos−1(x), and tan−1(x).
The input to the inverse trigonometric functions is a ratio, and the output is an angle. The function sin−1(x), for example, returns the angle whose sine ratio is x. The same goes for cos−1(x) and tan−1(x). We use the inverse trigonometric ratio to find an angle given the ratio of a pair of sides in a right triangle.
On the TI-84, press the 2ND button followed by either sin, cos, or tan to calculate sin−1, cos−1, or tan−1, respectively. Example 1 Using the accompanying figure, find the measure of angles M and N. Solution: Note that as expected, the measures found for ∠M and ∠N add to 90°. Example 2 Right triangle BTP has a right angle at T, BT = 4, and PT = 11. What is the value of cos(P)? Round to the nearest thousandth.
Solution: The method we saw previously was to use the Pythagorean theorem to find BP and then calculate cos(P) using .
Alternatively, we could use the inverse tangent function to find m∠P and then use the calculator to find cos(P). Example 3 In the accompanying figure, m∠GDO = 90°, m∠DSG = 90°, SG = 3, and SD = 5. Find the measure of angle O. Round to the nearest 0.01 degree. Solution: Strategy—Use the inverse tangent function to find m∠G. Then use the triangle angle sum theorem to find m∠O. Using Trigonometric Ratios to Find an Unknown Side Given any one angle in a triangle and any one side, all the remaining sides can be determined using trigonometric ratios. Write a ratio using a ratio of the known side to the unknown side. Set that ratio equal to the appropriate trigonometric ratio. Solve the resulting proportion for the unknown side, using your calculator to find the value of the trigonometric ratio. Example 1 Right triangle SOX has a right angle at O. The measure of angle S = 72°, and XO = 12. Find the length of OS to the nearest tenth. Solution: Strategy—Relative to ∠S, the given side and the unknown side are the opposite and adjacent, which suggests using the tangent ratio. Example 2 The perimeter of an equilateral triangle is 60 inches. What is the length of its altitude? Round your answer to the nearest 0.1 inch. Solution: Each side of the triangle measures inches, or 20 inches. By sketching the triangle, we see that relative to the 60° angle, the altitude is the opposite side and the 20-inch side is the hypotenuse. The opposite and hypotenuse form the sine ratio: Example 3 A cell phone tower is held in place with two support wires, as shown in the accompanying figure. The angle made by the wire and the ground is 55°. The distance between the base of the tower and the wire is 20 feet. How tall is the tower? What is the total length of wire used? Round all answers to the nearest tenth. Solution: Relative to the 55° angle, the 20-foot distance along the ground is the adjacent and the tower is the opposite. So we can use the tangent ratio. The tower is 28.6 feet high. Now that we have two sides of the triangle, we can use either the Pythagorean theorem or the cosine ratio to find the length of the wire. Continuing with the trigonometry approach, we see that the 20-foot distance is the adjacent and the wire is the hypotenuse. So we use the cosine ratio. There are two wires, so the total length of wire is (2) (34.8689) = 69.7 ft. Cofunctions The two acute angles in a right triangle must be complementary. As you may have noticed, the sine of one of the acute angles is equal to the cosine of other acute angle as shown in Figure 8.10. The sine and cosine are referred to as cofunctions. Cofunction relationship between sine and cosine Cofunction Relationship The sine of an angle is equal to the cosine of its complement: sin(A) = cos(90 − A) The cosine of an angle is equal to the sine of its complement: cos(A) = sin(90 − A) Example 1 Given right triangle ABC with right angle C, sin(A) is always equal to which of the following? sin(90 − A) cos(90 − A) cos(A) tan(B) Solution: The correct cofunction relationship is choice (2), cos(90 − A). Example 2 Given complementary angles M and N, if sin(M) = (a2 − 2) and cos(N) = a, find the value of a. Solution: Using the cofunction relationships, we know sin(M) = cos(N). By substituting, we find: The negative value of a would lead to a negative value for cosine. The cosine ratio of any acute angle is positive so we exclude the negative value. The value of a is 2. Math Fact In right triangle trigonometry, we deal with only acute positive angles. However, angles need not be acute or positive. In the study of trigonometry, the trigonometric ratios can be applied to both negative angles and obtuse angles. Angle of Elevation and Depression Many practical applications involve angles of elevation and angles of depression. Definitions Angle of elevation: The angle formed by the horizontal and the line directed upward to an object Angle of depression: The angle formed by the horizontal and the line directed downward to an object An angle of elevation involves a situation where someone or something is looking upward at another object and the horizontal is often the ground. A vertical line to complete the right triangle often has to be sketched. An angle of depression involves someone or something that is at a higher elevation looking down. The horizontal may also have to be sketched.
Figure illustrates an angle of elevation and an angle of depression.
Angle of elevation and angle of depression
Angle of elevation and angle of depression problems involve finding a missing angle or side using a trigonometric ratio.
The best strategy is to do the following: Start with a detailed sketch. Identify the triangles formed and relevant trigonometric ratios. Consider other parallel line and other relationships if you are still missing parts. Notice that the angle of elevation and the angle of depression shown in Figure 8.11 are congruent alternate interior angles. Consider working with more than one triangle, especially if the problem involves an object that moves from one position to another. Example 1 An airplane is flying level and is approaching a radar station. The radar station measures a straight-line distance to the airplane of 15,000 ft when the angle of elevation is 61°. What is the altitude of the airplane? What is the horizontal distance from the airplane to the radar station? A short time later, the angle of elevation to the plane is 72°. If the altitude of the airplane remains the same, what is the horizontal distance the airplane has traveled? Round answers to the nearest foot.
Solution: Sketch a vertical segment from the airplane to the ground to complete a right triangle. This distance, h, represents the elevation of the airplane and is opposite the angle of elevation. The 15,000 ft distance is the hypotenuse, so we should use the sine ratio. The airplane is flying at an altitude of 13,119 ft. The distance along the ground is the adjacent, so we use cosine. The plane is now closer to the radar station. The angle of elevation is 72° and the elevation is still 13,119 ft. The altitude is opposite the angle of elevation. We are looking for the horizontal distance, d, which is adjacent to the angle of elevation. The tangent ratio applies here.
Subtract to find the horizontal distance the plane has traveled. Example 2 Frank is standing at the top of a water tower and looks down at the town below. Frank measures an angle of depression of 52° to his house. He knows the water tower is 120 ft tall. How far along the ground is the tower from his house? Round to the nearest foot. Solution: Start with a sketch, and fill in the appropriate dimensions, horizontals, and verticals. The angle of depression is 52°. It is easier in the problem to find the angle complementary to the angle of depression and work with that triangle. The complementary angle is 38°. The height of the tower is adjacent to the 38° angle, and the distance along the ground is adjacent to the 38° angle. So we use the tangent ratio. The house is 94 feet from the water tower. Example 3 A lighthouse sits on a 100 ft cliff. An observer in a boat in the harbor notes an angle of elevation to the base of the cliff of 28° and an angle of elevation to the top of the lighthouse of 44°. What is the height of the lighthouse, rounded to the nearest foot? Solution: Strategy—First find the length of AB using ΔABC. Then find the length of BD using ΔABD. In ΔABC, BC is opposite ∠A and AB is the adjacent. So we use the tangent. In ΔABD, AB is the adjacent and BD is the opposite. So we use the tangent again.
The height of the lighthouse, DC, is found by subtracting the height of the cliff from BD.
The height of the lighthouse is 81.6 ft.
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