(B) Let the height of the pole $AB = h$. In $\Delta ABC$,$\tan 60^{\circ} = \frac{AB}{BC}$ $\Rightarrow \sqrt{3} = \frac{h}{BC}$ $\Rightarrow BC = \fr

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$\frac{7\sqrt{3}}{2}(\sqrt{3}+1)$

(B) Let the height of the pole $AB = h$. In $\Delta ABC$,$\tan 60^{\circ} = \frac{AB}{BC}$ $\Rightarrow \sqrt{3} = \frac{h}{BC}$ $\Rightarrow BC = \frac{h}{\sqrt{3}}$. In $\Delta ABD$,$\tan 45^{\circ} = \frac{AB}{BD}$ $\Rightarrow 1 = \frac{h}{BD}$ $\Rightarrow BD = h$. Since $BD = BC + CD$,we have $h = \frac{h}{\sqrt{3}} + 7$. $h - \frac{h}{\sqrt{3}} = 7 \Rightarrow h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = 7$. $h = \frac{7\sqrt{3}}{\sqrt{3}-1}$. Rationalizing the denominator: $h = \frac{7\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{7\sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{7\sqrt{3}}{2}(\sqrt{3}+1) \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

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$AB$ 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

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$\frac{7\sqrt{3}}{2(\sqrt{3}-1)}$
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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