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A view tower AB is 45 m high. From the top of the tower A, the angle of depression to the roof of a house C is 60 degrees. The distance from the base of the tower D to the house is 15 m. Find the heig

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2 parts
a

45−153 m45 - 15\sqrt{3} \text{ m}

19.02 m19.02 \text{ m}

b

arctan⁡(3)\arctan(3)

71.57∘71.57^\circ

Step-by-step solution

5 steps

Here's the solution to question 14:

Given:
Height of view tower AB = 45 m
Distance from tower to house BD = 15 m
Angle of depression from the top of the tower (A) to the roof of the house (C) = 60°

Let's denote the height of the house as CD.
Draw a horizontal line CE from the roof of the house (C) to the tower (AB), such that E is on AB.
Then, BCDE forms a rectangle.
Therefore, the horizontal distance EC = BD = 15 m.
The height of the house CD = BE.

Part a· 3 steps

Find the height of the house (CD).

  1. Identify the relevant triangle for the angle of depression.

    The angle of depression from A to C is 60°. This means that if we draw a horizontal line from A, the angle between this line and AC is 60°. Due to alternate interior angles, ∠ACE=60∘\angle ACE = 60^\circ.

  2. Use trigonometry in △AEC\triangle AEC.

    In the right-angled triangle AEC:

    tan⁡(∠ACE)=OppositeAdjacent=AEEC\tan(\angle ACE) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AE}{EC}

    We know ∠ACE=60∘\angle ACE = 60^\circ and EC=15 mEC = 15 \text{ m}.

    tan⁡(60∘)=AE15\tan(60^\circ) = \frac{AE}{15} 3=AE15\sqrt{3} = \frac{AE}{15} AE=153 mAE = 15\sqrt{3} \text{ m}

  3. Calculate the height of the house (CD).

    The total height of the tower AB = AE + EB.
    We know AB = 45 m and AE = 15315\sqrt{3} m.

    45=153+EB45 = 15\sqrt{3} + EB EB=45−153 mEB = 45 - 15\sqrt{3} \text{ m}

    Since CD = EB, the height of the house is:

    CD=45−153 mCD = 45 - 15\sqrt{3} \text{ m}

    Using 3≈1.732\sqrt{3} \approx 1.732:

    CD≈45−15(1.732)CD \approx 45 - 15(1.732) CD≈45−25.98CD \approx 45 - 25.98 CD≈19.02 mCD \approx 19.02 \text{ m}

    The height of the house is 45−153 m\boxed{45 - 15\sqrt{3} \text{ m}} or approximately 19.02 m\boxed{19.02 \text{ m}}.

Part b

arctan⁡(3)\arctan(3)

Find the angle of elevation from the base of the house (D) to the top of the tower (A).

  1. Identify the relevant triangle for the angle of elevation.

    We need to find ∠ADB\angle ADB. Consider the right-angled triangle ABD.

  2. Use trigonometry in △ABD\triangle ABD.

    In △ABD\triangle ABD:

    tan⁡(∠ADB)=OppositeAdjacent=ABBD\tan(\angle ADB) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BD}

    We know AB = 45 m and BD = 15 m.

    tan⁡(∠ADB)=4515\tan(\angle ADB) = \frac{45}{15} tan⁡(∠ADB)=3\tan(\angle ADB) = 3 ∠ADB=arctan⁡(3)\angle ADB = \arctan(3)

    Using a calculator, arctan⁡(3)≈71.565∘\arctan(3) \approx 71.565^\circ.
    The angle of elevation from the base of the house to the top of the tower is arctan⁡(3)\boxed{\arctan(3)} or approximately 71.57∘\boxed{71.57^\circ}.

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