Question

PQR is a triangular park where PQ = PR = 200 m. A TV tower is located in the center of the QR. If the vertical angles of the top of the tower at points P, Q and R are 45 °, 30 ° and 30° respectively, what is the height of the tower?

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Answer 1

Given :

• PQR is a triangular park where PQ = PR = 200 m.
• A TV tower is located in the center of the QR.
• The vertical angles of the top of the tower at points P, Q and R are 45 °, 30 ° and 30° respectively.

To find :

• Height of the tower.

Solution :

According to the attachment,

• Height of the tower = ST
• $\angle\:SPT=45^\degree$
• $\angle\:SQT=\angle\:SRT=30^\degree$

Let the height of the tower be x m.

In case of ∆SPT,

tan45° = ST/PT

→ 1 = x/PT

→ PT = x ...................(1)

In case of ∆STR,

tan30° = ST/TR

→ 1/√3 = x/TR

→ TR = x√3..............(2)

PQ = PR and ' T ' is the middle point of QR.

PT _|_ QR

∆PRT is a right triangle.

In case of ∆PRT,

PT² + TR² = PR²

→ x² +(√3x)² = 200²

→ x² + 3x² = 40000

→ 4x² = 40000

→ x = 100

Therefore, the height of the tower is 100 m.