If a line makes angles tan ^{-1} sqrt{7} and | vedclass

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(D) Let the angles made by the line with the $X$,$Y$,and $Z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively. Given: $\alpha = 	an ^{-1} \sqrt{7}

Vedclass · Accepted Answer

$\frac{\pi}{4}$ or $\frac{3 \pi}{4}$

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(D) Let the angles made by the line with the $X$,$Y$,and $Z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively. Given: $\alpha = 	an ^{-1} \sqrt{7} \implies 	an \alpha = \sqrt{7}$. Since $\sec^2 \alpha = 1 + 	an^2 \alpha = 1 + 7 = 8$,we have $\cos^2 \alpha = \frac{1}{8}$. $\beta = 	an ^{-1} \frac{\sqrt{5}}{\sqrt{3}} \implies 	an \beta = \frac{\sqrt{5}}{\sqrt{3}}$. Since $\sec^2 \beta = 1 + 	an^2 \beta = 1 + \frac{5}{3} = \frac{8}{3}$,we have $\cos^2 \beta = \frac{3}{8}$. We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Substituting the values: $\frac{1}{8} + \frac{3}{8} + \cos^2 \gamma = 1$ $\frac{4}{8} + \cos^2 \gamma = 1$ $\frac{1}{2} + \cos^2 \gamma = 1$ $\cos^2 \gamma = \frac{1}{2}$ $\cos \gamma = \pm \frac{1}{\sqrt{2}}$ Thus,$\gamma = \frac{\pi}{4}$ or $\gamma = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

If a line makes angles $\tan ^{-1} \sqrt{7}$ and $\tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}}$ with the $X$-axis and $Y$-axis respectively,then the angle made by it with the $Z$-axis is

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