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The angle of declination is 30° and Zama is standing on a 13 m high building. a) Calculate the distance between Zama and Peter. b) How far is Peter from the building.

Mathematics

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Answers

2 parts
a

26 m\text{26 m}

b

22.5 m\text{22.5 m}

Step-by-step solution

6 steps

Here's how to solve the problem:

Let's denote the position of Zama at the top of the building as A, the base of the building as B, and Peter's position on the ground as C. This forms a right-angled triangle ABC, with the right angle at B.
The height of the building (AB) is 13 m.
The angle of declination from Zama to Peter is 30∘30^\circ. This means the angle of elevation from Peter to Zama (∠ACB\angle ACB) is also 30∘30^\circ (alternate interior angles).

Part a· 3 steps

Calculate the distance between Zama and Peter.

  1. Identify the known values and the unknown.

    We know the opposite side (height of the building, AB=13AB = 13 m) and the angle (∠ACB=30∘\angle ACB = 30^\circ). We need to find the hypotenuse (distance between Zama and Peter, AC).

  2. Use the sine trigonometric ratio.

    sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} sin⁡(30∘)=ABAC\sin(30^\circ) = \frac{AB}{AC} sin⁡(30∘)=13AC\sin(30^\circ) = \frac{13}{AC}

  3. Solve for AC.

    AC=13sin⁡(30∘)AC = \frac{13}{\sin(30^\circ)}

    Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5:

    AC=130.5AC = \frac{13}{0.5} AC=26 mAC = 26 \text{ m}

    The distance between Zama and Peter is 26 m\boxed{\text{26 m}}.

Part b

22.5 m\text{22.5 m}

How far is Peter from the building?

  1. Identify the known values and the unknown.

    We know the opposite side (height of the building, AB=13AB = 13 m) and the angle (∠ACB=30∘\angle ACB = 30^\circ). We need to find the adjacent side (distance Peter is from the building, BC).

  2. Use the tangent trigonometric ratio.

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(30∘)=ABBC\tan(30^\circ) = \frac{AB}{BC} tan⁡(30∘)=13BC\tan(30^\circ) = \frac{13}{BC}

  3. Solve for BC.

    BC=13tan⁡(30∘)BC = \frac{13}{\tan(30^\circ)}

    Since tan⁡(30∘)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}:

    BC=131/3BC = \frac{13}{1/\sqrt{3}} BC=133 mBC = 13\sqrt{3} \text{ m} BC≈13×1.73205BC \approx 13 \times 1.73205 BC≈22.5 mBC \approx 22.5 \text{ m}

    Peter is approximately 22.5 m\boxed{\text{22.5 m}} from the building.

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