If a directed line makes angles 45^{circ} and | vedclass

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(B) Let the direction angles of the line be $\alpha = 45^{\circ}$,$\beta = 60^{\circ}$,and $\gamma = 	heta$.We know that the sum of the squares of th

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$120$

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(B) Let the direction angles of the line be $\alpha = 45^{\circ}$,$\beta = 60^{\circ}$,and $\gamma = 	heta$.We know that the sum of the squares of the direction cosines is $1$,i.e.,$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$.Substituting the given values: $\cos^{2}(45^{\circ}) + \cos^{2}(60^{\circ}) + \cos^{2}	heta = 1$.$(\frac{1}{\sqrt{2}})^{2} + (\frac{1}{2})^{2} + \cos^{2}	heta = 1$.$\frac{1}{2} + \frac{1}{4} + \cos^{2}	heta = 1$.$\frac{3}{4} + \cos^{2}	heta = 1$.$\cos^{2}	heta = 1 - \frac{3}{4} = \frac{1}{4}$.$\cos	heta = \pm \frac{1}{2}$.Since the angle $	heta$ is obtuse,$\cos	heta$ must be negative,so $\cos	heta = -\frac{1}{2}$.Therefore,$	heta = 120^{\circ}$.

If a directed line makes angles $45^{\circ}$ and $60^{\circ}$ with the $X$ and $Y$-axes respectively,then the obtuse angle $\theta$ made by the line with the $Z$-axis is (in $^{\circ}$)

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