First slide

Properties of triangles

Question

$If in a Δ A B C, b (b + c) = a^{2} and c (c + a) = b^{2}, then \cos A \cdot \cos B \cdot \cos C =$

Moderate

A

$\frac{1}{16}$
B

$\frac{1}{8}$
C

$- \frac{1}{8}$
D

$\frac{1}{2}$

Solution

$\begin{array}{l} Given that b (b + c) = a^{2} \\ \Rightarrow b c = a^{2} - b^{2} \\ \sin B \cdot \sin C = \sin^{2} A - \sin^{2} B \\ \sin B \cdot \sin C = \sin (A + B) \sin (A - B) \\ \sin B \cdot \sin C = \sin C \cdot \sin (A - B) \\ \sin B = \sin (A - B) \\ \Rightarrow 2 B = A \\ And similarly c (c + a) = b^{2} \\ \Rightarrow 2 C = B \\ We have A + B + C = π \\ \Rightarrow 2 B + B + \frac{B}{2} = π \Rightarrow \frac{7 B}{2} = π \Rightarrow B = \frac{2 π}{7} \\ \Rightarrow A = \frac{4 π}{7}, C = \frac{π}{7} \end{array}$
$∴ \cos \frac{π}{7} . \cos \frac{2 π}{7} . \cos \frac{4 π}{7} = \frac{\sin \frac{8 π}{7}}{8 \sin \frac{π}{7}} = - \frac{1}{8}$ $(\cos A \cos 2 A \cos 4 A . . . . . . . \cos 2^{n - 1} A = \frac{\sin 2^{n} A}{2^{n} \sin A})$

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