Question

If a straight line makes the angles ${60}^o,{45}^o$ and $\alpha$ with $X,Y$ and $Z$-axes respectively, then ${\sin }^2\alpha$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $1$
• E. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{3}{4}$

We know that, $l^2+m^2+n^2=1$ , where l, m, n are d.c's of line
$\Rightarrow {\cos }^2{60}^o+{\cos }^2{45}^o+{\cos }^2\alpha =1$
$\Rightarrow \frac{1}{4}+\frac{1}{2}=1-{\cos }^2\alpha$
$\Rightarrow {\sin }^2\alpha =\frac{3}{4}$