$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD = 7 \ m$. From $D$,the angle of elevation of the point $A$ is $45^{\circ}$. Then the height of the pole is
- A$\frac{7\sqrt{3}}{2(\sqrt{3}-1)}$
- B$\frac{7\sqrt{3}}{2}(\sqrt{3}+1)$
- C$\frac{7\sqrt{3}}{2}(\sqrt{3}-1)$
- D$\frac{7\sqrt{3}}{2(\sqrt{3}+1)}$
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The angle of elevation of the top $P$ of a tower from the feet of one person standing due South of the tower is $45^{\circ}$ and from the feet of another person standing due West of the tower is $30^{\circ}$. If the height of the tower is $5 \text{ m}$,then the distance (in meters) between the two persons is equal to $..........$.
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DifficultKVPY 2020
View SolutionThe angle of elevation of a tower from a point $A$ due south of it is $30^\circ$ and from a point $B$ due west of it is $45^\circ$. If the height of the tower is $100 \ m$,then $AB = \dots \ m$.
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View SolutionTwo vertical poles of equal heights are $120 \, m$ apart. On the line joining their bottoms,$A$ and $B$ are two points. The angle of elevation of the top of one pole from $A$ is $45^\circ$ and that of the other pole from $B$ is also $45^\circ$. If $AB = 30 \, m$,then the height of each pole is.....$m$.
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