Trigonometry & Inverse Trigonometry
Trigonometric equations in triangles
Grade 11
Question:
<p><strong>Ex. 11:</strong> In a triangle ABC, if \(4 \cos A \cos B + 4\sin A \sin B \sin C = 4\), then triangle ABC is</p>
<p>(a) right angle but not isosceles</p>
<p>(b) isosceles but not right angled</p>
<p>(c) right angle isosceles</p>
<p>(d) obtuse angled</p>
Step-by-Step Solution
Key Concept: Use the constraint $\cos(A-B) \leq 1$ to force equality conditions and deduce that both the triangle is isosceles and has a right angle.
<p><strong>Solution:</strong> We have $4 \cos A \cos B + 4\sin A \sin B \sin C = 4$</p><p>Dividing by 4: $\cos A \cos B + \sin A \sin B \sin C = 1$</p><p>$\cos(A-B) + \sin A \sin B \sin C = 1$</p><p>Since $\cos(A-B) \leq 1$, we need $\cos(A-B) = 1$ and $\sin A \sin B \sin C = 0$</p><p>From $\cos(A-B) = 1$, we get $A = B$</p><p>From $\sin A \sin B \sin C = 0$ in a triangle, we need $\sin C = 0$ is impossible, so $C = 90°$</p><p>Thus $A = B = 45°$ each, and $C = 90°$</p><p>∴ Triangle ABC is right angle isosceles.</p>
Correct Answer: C