Two towers face each other separated by a distance = 15 m. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. What is the height (in meters) of the second tower?

Height of second tower = 17.32m

Step-by-step explanation:

I have attached a diagram depicting the question.

From the diagram, The first tower is depicted by side AEB and the second tower CD.

While d is the distance that separates the two towers and it's given as 15m.

Now, since the angle of depression of the second tower’s base is 60°, then for triangle BAC. Angle C = 60°.

Thus; using trigonometric ratios;

tan 60° = AB/AC.

This gives; AB = d*tan 60°

Similarly, for the triangle BED, BE = d*tan 30°

Since, AE = CD, thus ;

CD = AB − BE

CD = d (tan 60° − tan 30°)

CD = 15(1.7321 − 0.5774)

CD = 15 × 1.1547

CD ≈ 17.32 m.

So, height of second tower = 17.32 m

its 4 because 2 times 2 is 4, if you double the number two you get four

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