If the angles made by a straight line with the coordinate axes are $\alpha, \frac{\pi}{2}-\alpha, \beta$,…
- 0
- $\frac{\pi}{6}$
- $\frac{\pi}{2}$
- $\pi$
Solution

But, given $\alpha=\alpha, \beta=\frac{\pi}{2}-\alpha, \gamma=\beta$ From Eq. (i) $\cos ^2 \alpha+\cos ^2\left(\frac{\pi}{2}-\alpha\right)+\cos ^2 \beta=1$ $\begin{array}{cc}\Rightarrow & \left(\cos ^2 \alpha+\sin ^2 \alpha\right)+\cos ^2 \beta=1 \\ \Rightarrow & \cos ^2 \beta=0 \\ \Rightarrow & \cos \beta=0=\cos \frac{\pi}{2} \\ \Rightarrow & \beta=\frac{\pi}{2}\end{array}$
Asked in: AP EAMCET 2011
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