Question

# The value of $$'\alpha '$$ so that $$sin^{-1} \frac{2}{\sqrt{5}}, sin^{-1} \frac{3}{\sqrt{10}}, sin^{-1} (\alpha )$$ are the angles of a triangle is

A
12
B
12
C
12
D
13

Solution

## The correct option is B $$\dfrac{1}{\sqrt{2}}$$$$sin^{-1} \frac{2}{\sqrt{5}} + sin^{-1} \frac{3}{\sqrt{10}} + sin^{-1} \alpha =\pi$$$$\Rightarrow tan^{-1}2 + tan^{-1}3+tan^{-1} \frac{\alpha }{\sqrt{1-\alpha ^{2}}}=\pi$$$$\Rightarrow \frac{\alpha }{\sqrt{1-\alpha ^{2}}}=1$$$$\alpha =\pm \frac{1}{\sqrt{2}}\Rightarrow \alpha =\frac{1}{\sqrt{2}}$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More