**1.** In right triangle ABC,

Hence, the height of the pole is 10 m.

**2. ** In right triangle ABC,

**3. ** In right triangle ABC,

Hence, the lengths of the slides are 3 m and 2√3 m respectively.

**4. ** In right triangle ABC,

**5. ** In right triangle ABC,

Hence the length of the string is 40√3 m.

**6.** AB = 30 m and PR = 1.5 m

AC = AB – BC

= AB – PR

= 30 – 1.5

= 28.5 m

In right triangle ACQ,

Hence, the walked towards the building is 19√3 m.

**7.** Let the height of the tower be *h *m. Then, in right triangle CBP,

Putting this value in eq. (i), we get,

∴** **The height of the tower is 20(√5 – 1) m. 8.

**8.** Let the height of the pedestal be *h *m.

∴ BC = *h *m

In right triangle ACP,

⇒ *h *= 0.8 (√3 + 1) m

Hence, the height of the pedestal is 0.8 (√3 + 1) m.

**9. ** Let the height of the building be *h *m.

**10.** In right triangle PRQ,

Also, BR = 80 – *h* *= *80 – 20 = 60 m

Hence the heights of the poles are 20√3 m each and the distances of the point from poles are 20 m and 60 m respectively.

**11. ** In right triangle ABC,

Hence height of the tower is 10 √5 m and the width of the canal is 10 m.

**12.** In right triangle ABD,

**13.** In right triangle ABQ,

Hence the distance between the two ships is 75 (√3 – 1) m. 14.

**14.** In right triangle ABC,

Hence the distance travelled by the balloon during the interval is

**15.** In right triangle ABP,

∵ Time taken by the car to travel a distance PQ = 6 seconds.

∵ Time taken by the car to travel a distance BQ, i.e. seconds.

Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.

**16. ** Let ∠ APB = θ

Then, ∠ AQB = (90° — θ*)*

[∠ APB and ∠ AQB are complementary]

In right triangle ABP,

⇒ AB = 6 m

Hence, the height of the tower is 6 m.

Proved.