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

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.

think the answer is 30 degrees. if not than it is 20 degrees. not 100% sure. let me know if it was right.

gill russ