How do I choose between using the Sine Rule and the Cosine Rule in a problem?

Use the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ when given two angles and any side, or two sides and an angle opposite to one of them. Use the Cosine Rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ when all three side lengths are given, or when two sides and their included angle are known.

What is the best strategy to tackle mixed questions with $r$, $R$, and $r_1, r_2, r_3$?

Convert all radii into half-angle trigonometric form using $r = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)$ and $r_1 = 4R \sin\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right)$. This transforms geometric relations into standard trigonometric identities that simplify rapidly.

Properties of Triangles Notes, Formulas, Traps & PYQs | ExamiFly

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024 (Foundation Standard)

In a triangle $ABC$ , if the sides are $a = 3$ , $b = 5$ , and $c = 7$ , then the value of $\cos C$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply the Cosine Rule for angle $C$ : $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ .

Step 2

Substitute the given values $a = 3$ , $b = 5$ , and $c = 7$ : $\cos C = \frac{3^2 + 5^2 - 7^2}{2(3)(5)} = \frac{9 + 25 - 49}{30}$ .

Step 3

Simplify the numerator: $\cos C = \frac{-15}{30} = -\frac{1}{2}$ , which corresponds to $C = 120^\circ$ .

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Question #2JEE Main 2023

In a triangle $ABC$ , if the side lengths are in the ratio $a : b : c = 4 : 5 : 6$ , then the ratio of the circumradius $R$ to the inradius $r$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Set sides $a = 4k$ , $b = 5k$ , $c = 6k$ . The semi-perimeter is $s = \frac{4k + 5k + 6k}{2} = \frac{15k}{2}$ .

Step 2

Calculate area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \frac{15\sqrt{7}k^2}{4}$ .

Step 3

Find $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \left(\frac{15\sqrt{7}k^2}{4}\right)} = \frac{8k}{\sqrt{7}}$ and $r = \frac{\Delta}{s} = \frac{\frac{15\sqrt{7}k^2}{4}}{\frac{15k}{2}} = \frac{\sqrt{7}k}{2}$ . Thus, $\frac{R}{r} = \frac{\frac{8k}{\sqrt{7}}}{\frac{\sqrt{7}k}{2}} = \frac{16}{7}$ .

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Mastering Properties of Triangles: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Sine Rule and Circumradius

Cosine Rule

Napier's Analogy (Tangent Rule)

Area of Triangle (Heron's & Radius Relations)

Ex-radii Formulas

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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