The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $ an ^{-1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance $40 \mathrm{~m}$ from the foot.

Question

The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $	an ^{-1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance $40 \mathrm{~m}$ from the foot.

Aicharya · Accepted Answer

$	an \beta=\frac{	an 	heta-	an \alpha}{1+	an 	heta \cdot 	an \alpha}$ or $\frac{3}{5}=\frac{\frac{\mathrm{h}}{40}-\frac{\mathrm{h}}{160}}{1+\frac{\mathrm{h}}{40} \cdot \frac{\mathrm{h}}{160}}$ $\mathrm{h}^2-200 \mathrm{~h}+6400=0, \mathrm{~h}=40$ or 160 metre Therefore possible height $=40$ metre

The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $\tan ^{-1} \frac{3}{5}$ at a point…

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