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(B) The direction cosines of a line are $\cos \alpha, \cos \beta, \cos \gamma$. We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given

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$90$

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(B) The direction cosines of a line are $\cos \alpha, \cos \beta, \cos \gamma$. We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given that $\alpha + \beta = 90^o$,we have $\beta = 90^o - \alpha$. Substituting this into the identity: $\cos^2 \alpha + \cos^2(90^o - \alpha) + \cos^2 \gamma = 1$ $\cos^2 \alpha + \sin^2 \alpha + \cos^2 \gamma = 1$ Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we get: $1 + \cos^2 \gamma = 1$ $\cos^2 \gamma = 0$ $\cos \gamma = 0$ Therefore,$\gamma = 90^o$.

$A$ line makes angles $\alpha, \beta, \gamma$ with the coordinate axes. If $\alpha + \beta = 90^o$,then $\gamma = $ ......... $^o$

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