Two posts are 2 text{ m} apart. Both posts ar | vedclass

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(B) Let $AD$ be the height of the tree,$h$. Let $C$ and $B$ be the positions of the two posts on the ground such that $CB = 2 	ext{ m}$. Let $DC = x

Vedclass · Accepted Answer

$(3+\sqrt{3}) 	ext{ m}$

Vedclass · Answer

(B) Let $AD$ be the height of the tree,$h$. Let $C$ and $B$ be the positions of the two posts on the ground such that $CB = 2 	ext{ m}$. Let $DC = x 	ext{ m}$.In $\Delta ADC$,the angle of elevation is $60^{\circ}$:$	an 60^{\circ} = \frac{AD}{DC} = \frac{h}{x}$$\sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}} \quad \dots(i)$In $\Delta ADB$,the angle of elevation is $45^{\circ}$:$	an 45^{\circ} = \frac{AD}{DB} = \frac{h}{x+2}$$1 = \frac{h}{x+2} \implies x+2 = h \implies x = h-2 \quad \dots(ii)$Equating $(i)$ and $(ii)$:$\frac{h}{\sqrt{3}} = h-2$$2 = h - \frac{h}{\sqrt{3}} = h \left(1 - \frac{1}{\sqrt{3}}ight) = h \left(\frac{\sqrt{3}-1}{\sqrt{3}}ight)$$h = \frac{2\sqrt{3}}{\sqrt{3}-1}$Rationalizing the denominator:$h = \frac{2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(3+\sqrt{3})}{3-1} = \frac{2(3+\sqrt{3})}{2} = (3+\sqrt{3}) 	ext{ m}$.

Two posts are $2 \text{ m}$ apart. Both posts are on the same side of a tree. If the angles of depression of these posts when observed from the top of the tree are $45^{\circ}$ and $60^{\circ}$ respectively,then the height of the tree is:

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