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(B) We know that the sum of the squares of the direction cosines of a line is $1$,i.e.,$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$. Given

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$90$

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(B) We know that the sum of the squares of the direction cosines of a line is $1$,i.e.,$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$. Given that $\alpha + \beta = 90^{\circ}$,we have $\alpha = 90^{\circ} - \beta$. Substituting this into the expression for $\cos \alpha$,we get $\cos \alpha = \cos(90^{\circ} - \beta) = \sin \beta$. Therefore,$\cos^{2} \alpha = \sin^{2} \beta$. Using the identity $\sin^{2} \beta = 1 - \cos^{2} \beta$,we get $\cos^{2} \alpha = 1 - \cos^{2} \beta$,which implies $\cos^{2} \alpha + \cos^{2} \beta = 1$. Substituting this into the original equation: $(1) + \cos^{2} \gamma = 1$. This simplifies to $\cos^{2} \gamma = 0$,which means $\cos \gamma = 0$. Thus,$\gamma = 90^{\circ}$.

$A$ line makes angles $\alpha, \beta, \gamma$ with the coordinate axes and $\alpha+\beta=90^{\circ}$,then $\gamma=$ (in $^{\circ}$)

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