A flag pole and a building stand on the same horizontal level. From the point p at the bottom of a building the angle of the top T of the flagpole is 65°. From the point Q of the building the angle of elevation of the point T is 25°. If the building is 20m high. calvulate the distance PT, The heigt of the flagpole,The distance QT

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oeerivona oeerivona · Answer 1 · 2021-05-27T02:31:12+02:00

Answer:

PT is approximately 28.2 m

The height of the flagpole is approximately 25.56 m

QT is approximately 13.15 m

Step-by-step explanation:

Part A

From the attached drawing of the building and the flagpole, we have;

∠TQP = 25° + 90° = 115°

∠TPQ = 90° - 65° = 25°

∴ ∠QTP = 180° - (115° + 25°) = 40°

By sine rule, we have;

[tex]\dfrac{\overline {QT}}{sin(\angle TPQ)} = \dfrac{\overline {PT}}{sin(\angle TQP)} = \dfrac{\overline {QP}}{sin(\angle QTP)}[/tex]

[tex]\therefore \overline {PT} = sin(\angle TQP) \times \dfrac{\overline {QP}}{sin(\angle QTP)}[/tex]

∴ PT = sin(115°) × 20 m/(sin(40°)) = 28.1992924 ≈ 28.2

PT ≈ 28.2 m

Part B

The height of the flagpole, Ta = PT × sin(∠TPa)

∴ Ta ≈ 28.2 × sin(65°) ≈ 25.56

The height of the flagpole, Ta ≈ 25.56 m

Part C

[tex]\overline {QT} = sin(\angle TPQ) \times \dfrac{\overline {QP}}{sin(\angle QTP)}[/tex]

The distance QT= sin(25°) × 20 m/(sin(40°)) ≈ 13.15

QT ≈ 13.15 m