How to Solve: Solution of Triangles (Sine Rule, Cosine Rule, Projection Formula, Inradius/Circumradius) – IIT JEE Guide

How to Solve: Solution of Triangles (Sine Rule, Cosine Rule, Projection Formula, Inradius/Circumradius) – IIT JEE Guide

Introduction

Mastering Solution of Triangles unlocks 10-15 marks in IIT JEE (Main + Advanced) every year—enough to push you into the top 1%. Whether it’s finding a missing side in a non-right triangle, calculating the radius of a circumscribed circle, or solving real-world navigation problems, these formulas are your secret weapon for geometry questions that others skip.

What You Need To Know First

Before diving in, ensure you’re 100% clear on: 1. Basic trigonometric ratios (sin, cos, tan) – You must know these for any angle, not just 30°/45°/60°. 2. Pythagoras’ theorem – Even though we’re dealing with non-right triangles, right-triangle intuition helps. 3. Angle sum property of triangles (A + B + C = 180°) – This is non-negotiable for solving triangles.

If any of these feel shaky, stop now and review them—this topic builds on them.

Key Vocabulary

Formulas To Know

Memorize these—IIT JEE does NOT provide them in the formula sheet!

1. Sine Rule (Law of Sines)

Formula: [ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R ] Variables: - (a, b, c) = sides opposite angles (A, B, C) respectively. - (R) = circumradius.

When to use: - When you have 2 angles and 1 side (AAS/ASA). - When you have 2 sides and a non-included angle (SSA)—but watch for the ambiguous case!

2. Cosine Rule (Law of Cosines)

Formula: [ a^2 = b^2 + c^2 - 2bc \cos A ] [ b^2 = a^2 + c^2 - 2ac \cos B ] [ c^2 = a^2 + b^2 - 2ab \cos C ]

Variables: - (a, b, c) = sides opposite angles (A, B, C).

When to use: - When you have 3 sides (SSS) and need an angle. - When you have 2 sides and the included angle (SAS) and need the third side.

3. Projection Formula

Formula: [ a = b \cos C + c \cos B ] [ b = a \cos C + c \cos A ] [ c = a \cos B + b \cos A ]

Variables: - (a, b, c) = sides opposite angles (A, B, C).

When to use: - When you need to express a side in terms of angles. - Useful in trigonometric identities and optimization problems.

4. Area of a Triangle (Trigonometric Form)

Formula: [ \text{Area} = \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B ]

When to use:
- When you have 2 sides and the included angle (SAS). - Faster than Heron’s formula in most cases.

5. Inradius (r) and Circumradius (R)

Formulas: [ r = \frac{\text{Area}}{s} \quad \text{(where } s = \frac{a+b+c}{2} \text{ is the semi-perimeter)} ] [ R = \frac{abc}{4 \times \text{Area}} ]

When to use: - Inradius (r): When you need the radius of the inscribed circle. - Circumradius (R): When you need the radius of the circumscribed circle.

Step-by-Step Method

Follow these exact steps for any triangle problem:

Step 1: Draw the Triangle & Label Everything

• Sketch the triangle.
• Label all given sides and angles (use (a, b, c) for sides opposite (A, B, C)).
• Mark unknowns with question marks.

Step 2: Check What’s Given

• AAS/ASA? → Use Sine Rule.
• SAS? → Use Cosine Rule (for side) or Area Formula (for area).
• SSS? → Use Cosine Rule (for angles).
• SSA? → Use Sine Rule, but check for ambiguous case (two possible triangles).

Step 3: Apply the Correct Formula

• Write the formula before plugging in numbers.
• Solve for one unknown at a time.

Step 4: Verify the Answer

• Check angle sum: (A + B + C = 180°).
• Check side lengths: The largest side should be opposite the largest angle.
• Use another formula (e.g., if you used Sine Rule, cross-check with Cosine Rule).

Step 5: Handle Special Cases

• Ambiguous case (SSA): If (\sin \theta > 1), no solution. If (\sin \theta = 1), one right triangle. If (0 < \sin \theta < 1), two possible triangles.
• Right triangle? Simplify using Pythagoras or basic trig ratios.

Worked Examples

Example 1 – Basic (Sine Rule)

Problem: In (\triangle ABC), (A = 30°), (B = 45°), and (a = 8). Find side (b).

Solution: 1. Draw & label:
- (A = 30°), (B = 45°), (a = 8) (opposite (A)).
- (b = ?) (opposite (B)).

1. Check given: AAS → Sine Rule.
2. Apply Sine Rule:
   [
   \frac{a}{\sin A} = \frac{b}{\sin B}
   ]
   [
   \frac{8}{\sin 30°} = \frac{b}{\sin 45°}
   ]
3. Plug in values:
   [
   \frac{8}{0.5} = \frac{b}{\frac{\sqrt{2}}{2}} \implies 16 = \frac{b \times 2}{\sqrt{2}}
   ]
   [
   b = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}
   ]
4. Verify:
5. (C = 180° - 30° - 45° = 105°) (valid).
6. Largest side should be opposite largest angle ((C = 105°)).

Answer: (b = 8 \sqrt{2})

What we did and why: - Used Sine Rule because we had two angles and one side (AAS). - No ambiguous case here because we had two angles (AAS is always unique).

Example 2 – Medium (Cosine Rule + Area)

Problem: In (\triangle ABC), (a = 7), (b = 5), (C = 60°). Find: (i) Side (c) (ii) Area of the triangle

Solution: (i) Find side (c): 1. Given: SAS → Cosine Rule. 2. Apply Cosine Rule:
[
c^2 = a^2 + b^2 - 2ab \cos C
]
[
c^2 = 7^2 + 5^2 - 2 \times 7 \times 5 \times \cos 60°
]
[
c^2 = 49 + 25 - 70 \times 0.5 = 74 - 35 = 39
]
[
c = \sqrt{39}
]

(ii) Find area: 1. Given: SAS → Area formula. 2. Apply:
[
\text{Area} = \frac{1}{2} ab \sin C = \frac{1}{2} \times 7 \times 5 \times \sin 60°
]
[
\text{Area} = \frac{35}{2} \times \frac{\sqrt{3}}{2} = \frac{35 \sqrt{3}}{4}
]

Answer: (i) (c = \sqrt{39}) (ii) Area = (\frac{35 \sqrt{3}}{4})

What we did and why: - Used Cosine Rule for SAS to find the missing side. - Used Area formula because we had two sides and the included angle. - No need for Heron’s formula—this was faster.

Example 3 – Exam-Style (Ambiguous Case + Inradius)

Problem: In (\triangle ABC), (a = 5), (b = 4), (A = 30°). Find: (i) All possible values of angle (B). (ii) The inradius (r) of the triangle(s).

Solution: (i) Find angle (B): 1. Given: SSA → Sine Rule (but ambiguous case). 2. Apply Sine Rule:
[
\frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{5}{\sin 30°} = \frac{4}{\sin B}
]
[
\frac{5}{0.5} = \frac{4}{\sin B} \implies 10 = \frac{4}{\sin B} \implies \sin B = 0.4
] 3. Find possible angles:
- (B = \sin^{-1}(0.4) \approx 23.58°)
- OR (B = 180° - 23.58° = 156.42°) 4. Check validity:
- If (B = 156.42°), then (A + B = 30° + 156.42° = 186.42° > 180°) → Invalid.
- Only valid solution: (B \approx 23.58°).

But wait! Let’s verify: - If (B = 23.58°), then (C = 180° - 30° - 23.58° = 126.42°). - Now, check if two triangles are possible: - Case 1: (B = 23.58°), (C = 126.42°). - Case 2: (B = 156.42°) → Invalid (as above). - Conclusion: Only one triangle is possible.

(ii) Find inradius (r): 1. Find side (c) (using Sine Rule):
[
\frac{c}{\sin C} = \frac{a}{\sin A} \implies c = \frac{a \sin C}{\sin A} = \frac{5 \times \sin 126.42°}{0.5}
]
[
\sin 126.42° = \sin (180° - 53.58°) = \sin 53.58° \approx 0.8
]
[
c = \frac{5 \times 0.8}{0.5} = 8
] 2. Find semi-perimeter (s):
[
s = \frac{a + b + c}{2} = \frac{5 + 4 + 8}{2} = 8.5
] 3. Find area (using Heron’s formula):
[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{8.5 \times 3.5 \times 4.5 \times 0.5} \approx 9.92
] 4. Find inradius (r):
[
r = \frac{\text{Area}}{s} = \frac{9.92}{8.5} \approx 1.17
]

Answer: (i) (B \approx 23.58°) (only valid solution) (ii) (r \approx 1.17)

What we did and why: - SSA case → Sine Rule but checked for ambiguity. - Only one valid triangle existed (the other angle sum exceeded 180°). - Used Heron’s formula for area because we had all three sides. - Inradius formula requires area and semi-perimeter.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second crash course for Solution of Triangles in IIT JEE:

1. Sine Rule: (\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R). Use when you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA)—but watch for the ambiguous case!
2. Cosine Rule: (a^2 = b^2 + c^2 - 2bc \cos A). Use for SAS (find side) or SSS (find angle).
3. Area: (\frac{1}{2} ab \sin C)—faster than Heron’s if you have SAS.
4. Inradius (r): (\frac{\text{Area}}{s}). Circumradius (R): (\frac{abc}{4 \times \text{Area}}).
5. Ambiguous case? If SSA, check if (\sin \theta > 1) (no solution) or two angles possible.
6. Always draw the triangle first—label everything. Verify angle sum = 180° and largest side opposite largest angle.

You’ve got this. Now go crush those 10-15 marks!