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,

= 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.