Question

A Surveyar want to find out the height of a
tower. He measure angle of A as tan A=3/4
What is the height of the tower if A is 40 m from its base as in the figure?​

Answer 1

Answer:

Given:-

• ∠ A = tan A = $\dfrac{3}{4}$

• AB = 40m

To Find:-

• Height of Tower (BC)

Solution:-

As we know,

$tan \:θ = \dfrac{opp.}{adj.}$

Then,

⟾ $tan \: A = \: \dfrac{BC}{AB}$

⟾ $\dfrac{3}{4} \:= \: \dfrac{BC}{AB}$

⟾ $\dfrac{3}{4} \:= \: \dfrac{BC}{40}$

⟾ $BC \:= \: \dfrac{40×3}{4}$

⟾ $BC \:= \: \dfrac{120}{4}$

⟾ $BC \:= \: 30\: cm$

Hence, The Height of Tower (BC) is 30 cm.

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More to know:-

• $sin \:θ = \dfrac{opp.}{hyp.}$

• $cos \:θ = \dfrac{adj.}{hyp.}$

• $cosec \:θ = \dfrac{hyp.}{opp.}$

• $sec \:θ = \dfrac{hyp.}{adj.}$

• $cot \:θ = \dfrac{adj.}{opp.}$

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