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(B) For a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines satisfy the relation: $\cos^2 \alpha + \cos^2 \bet

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$90$

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(B) For a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines satisfy the relation: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given that $\alpha + \beta = 90^\circ$,we have $\beta = 90^\circ - \alpha$. Substituting this into the relation: $\cos^2 \alpha + \cos^2(90^\circ - \alpha) + \cos^2 \gamma = 1$. Since $\cos(90^\circ - \alpha) = \sin \alpha$,the equation becomes: $\cos^2 \alpha + \sin^2 \alpha + \cos^2 \gamma = 1$. Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get: $1 + \cos^2 \gamma = 1$. This implies $\cos^2 \gamma = 0$,so $\cos \gamma = 0$. Therefore,$\gamma = 90^\circ$.

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

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