Question

7. From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.
Find the height of the tower.
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Answer 1

Answer:

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Answer 2

Answer:

$\bigstar{\bold{Height\:of\:tower\:=\:20(\sqrt{3}\:-\:1)\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Height of the building (BD) = 20 m
• Angle of elevation at the bottom = 45°
• Angle of elevation at the top = 60°

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Height of the tower (AD)

$\Large{\underline{\underline{\bf{Solution:}}}}$

➣ Let x = AD be the height of the tower

➣ First we need to find BC

➣ Consider ΔCBD

    tan 45 = oppposite/adjacent

    tan 45 = BD/BC

        1 = 20/BC

     BC = BD = 20 m

➣ Consider ΔCBA

    tan 60 = AD + DB/BC

    tan 60 = 20 + x /20

     √3 = 20+x/20

    20√3 = 20 + x

    20√3 - 20 = x

    x = 20(√3 - 1) m

 $\boxed{\bold{Height\:of\:tower\:=\:20(\sqrt{3}\:-\:1)\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

• Sin A  = opposite/hypotenuse
• Cos A  = adjacent /hypotenuse
• Tan A = opposite / adjacent