First slide

Multiple and sub- multiple Angles

Question

In $Δ A B C \tan A + \tan B + \tan C = 6 a n d \tan A \cdot \tan B = 2 then the possible value of \sin^{2} A : \sin^{2} B : \sin^{2} C =$

Moderate

A

$8 : 9 : 5$
B

$8 : 5 : 9$
C

$5 : 9 : 8$
D

$5 : 8 : 7$

Solution

$\begin{array}{l} \tan A + \tan B + \tan C = \tan A \tan B \tan C = 6 \\ \Rightarrow 2 \tan C = 6 (∵ \tan A \tan B = 2) \\ \Rightarrow \tan C = 3 \\ ∴ \tan A + \tan B = 6 - 3 \\ \Rightarrow \tan A + \tan B = 3 \\ \Rightarrow \tan A = 2 a n d \tan B = 1 \\ \Rightarrow \sin^{2} A : \sin^{2} B : \sin^{2} C = \frac{\tan^{2} A}{1 + \tan^{2} A} : \frac{\tan^{2} B}{1 + \tan^{2} B} : \frac{\tan^{2} C}{1 + \tan^{2} C} \\ = \frac{4}{5} : \frac{1}{2} : \frac{9}{10} \\ = 8 : 5 : 9 \end{array}$

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