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Question 51

The distance between the tops of two trees 20 m and 28 m high is 17 m. The horizontal distance between the two trees is

Solution

AD and CE are two trees of height 28 m and 20 m respectively. AC = 17 m

In right $$\triangle$$ ABC,

=> $$(BC)^2=(AC)^2-(AB)^2$$

=> $$(BC)^2=(17)^2-(8)^2$$

=> $$(BC)^2=289-64=225$$

=> $$BC=\sqrt{225}=15$$ m

=> Ans - (C)

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