First slide

Compound angles

Question

In a $∆$ ABC, if tan A : tan B : tan c = 3 : 4: 5, then the value of sin A . sin B . sin C is equal to

Moderate

A

$\frac{2}{\sqrt{5}}$
B

$\frac{2 \sqrt{5}}{7}$
C

$\frac{2 \sqrt{5}}{9}$
D

$\frac{2}{3 \sqrt{5}}$

Solution

Let tan A =3k, tan B = 4k and tan C = 5k
intriangle ABC,A+B+C=180
tan A+ tan B + tan C= tan A tan B tan C,
$\begin{array}{l} \Rightarrow 12 k = 60 k^{3} \\ \Rightarrow k = \frac{1}{\sqrt{5}} \end{array}$
$∴ \tan A = \frac{3}{\sqrt{5}}, \tan B = \frac{4}{\sqrt{5}}$
$∴ \sin A = \frac{3}{\sqrt{14}}, \sin B = \frac{4}{\sqrt{21}}, \sin C = \frac{\sqrt{5}}{\sqrt{6}}$
Hence $\sin A \cdot \sin B \cdot \sin C = \frac{2 \sqrt{5}}{7}$

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