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Question

A statue, $$1.6$$ m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $$\displaystyle { 60 }^{ \circ  }$$ and from the same point the angle of elevation of the top of the pedestal is $$\displaystyle { 45 }^{ \circ  }$$. Find the height of the pedestal.


Solution


Let $$AB$$ be the length of the statue and $$BC$$ be the length of the pedestal.

In $$\triangle BCD$$, 

$$\dfrac{BC}{CD} = \tan 45^o$$ 

$$\therefore BC=CD$$

In $$\triangle ACD$$, 
$$\tan 60^o=\dfrac {AC}{CD}$$

$$\dfrac{AB+BC}{CD}=\sqrt{3}$$

$$BC+1.6=\sqrt{3}CD$$
$$BC+1.6=\sqrt{3}BC$$

$$BC=\dfrac{1.6}{\sqrt{3}-1}$$

$$BC=\dfrac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$

$$=\dfrac{1.6(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2}=\dfrac{1.6(\sqrt{3}+1)}{2}$$

$$=0.8(\sqrt{3}+1)$$ m

495460_466367_ans_360ba8d165b448d5a09caf1b79b15ce7.png

Mathematics
RS Agarwal
Standard X

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