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A line makes angles $\propto, \beta, \gamma$ with the co-ordinate axes and $\propto+\beta=90^{\circ}$, then $\gamma=$
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$90^{\circ}$
We know that $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
It is given that $\alpha+\beta=90^{\circ} \Rightarrow \alpha=90^{\circ}-\beta \Rightarrow \cos \alpha=\cos \left(90^{\circ}-\beta\right)$
$\therefore \cos \alpha=\sin \beta \Rightarrow \cos ^{2} \alpha=\sin ^{2} \beta=1-\cos ^{2} \beta \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta=1$
Thus $1+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=0 \Rightarrow \gamma=90^{\circ}$
It is given that $\alpha+\beta=90^{\circ} \Rightarrow \alpha=90^{\circ}-\beta \Rightarrow \cos \alpha=\cos \left(90^{\circ}-\beta\right)$
$\therefore \cos \alpha=\sin \beta \Rightarrow \cos ^{2} \alpha=\sin ^{2} \beta=1-\cos ^{2} \beta \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta=1$
Thus $1+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=0 \Rightarrow \gamma=90^{\circ}$
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