**Q) From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between tower and the building.**

**Ans: **

Let’s start with the diagram for this question:

Here we have tower , PQ is the 15m high building and AB is the tower.

Angles of elevation from P & Q to A are given.

We need to find Height H and distance D.

Let’s make a simplified diagram for the question.

**Step 1: Horizontal distance between building & tower:**

In Δ ABQ, tan Q = tan 60° =

H = D **√**3 ………………….. (i)

**Step 2: **In Rectangle PQBC, PQ = CB = 15

Next, AC = AB – CB = H – 15

Next, In Δ ACP, cos P = tan 30° =

D = √3 (H – 15) ……………………(ii)

**Step 3:** Let’s substitute value of H from equation (i) in equation (ii), we get:

D = **√**3 (H – 15)

∴ D = **√**3 (D**√**3 – 15) = 3D – 15 **√**3

∴ 3D – D = 15 **√**3

∴ D = = 12.975 m

**Therefore the distance between tower and the building is 12.975 m**

**Step 4:** We substitute value of D = in equation (i), we get:

H = D **√**3 = **√**3

∴ H = = 22.5 m

**Therefore, the height of the tower is 22.5 m.**

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