# HEIGHTS AND DISTANCES

Ans .

$$60^{\circ}$$

1. Explanation :

Let AB be the pole and AC be its shadow.

Let angle of elevation,$$\angle$$ACB=$$\theta$$

Then, AB = 2$$\sqrt{3}$$m AC = 2 m.

Tan $$\theta$$=AB/AC = $$\sqrt[2]{\frac{3}{2}}$$=$$\sqrt{3}$$=>$$\theta$$

So, the angle of elevation is $$60^{\circ}$$

Ans .

9.5

1. Explanation :

Let AB be the wall and BC be the ladder.

Then, $$\angle$$ACB= $$60^{\circ}$$ and BC = 19 m.

Let AC = x metres

AC/BC = cos $$60^{\circ}$$ => x/19 = 1/2=>x=19/2 = 9.5

Distance of the foot of the ladder from the wall = 9.5

Ans .

20.76 m

1. Explanation :

Let AB be the tower and C and D be the points of observation. Then,

AB/AD = tan $$60^{\circ}$$ = $$\sqrt{3}$$ => AD = AB/$$\sqrt{3}$$= h/$$\sqrt{3}$$

AB/AC = tan $$30^{\circ}$$ = 1/$$\sqrt{3}$$ AC=AB x$$\sqrt{3}$$ = h$$\sqrt{3}$$

CD = (AC-AD) = (h$$\sqrt{3}$$-h/$$\sqrt{3}$$)

(h$$\sqrt{3}$$-h/$$\sqrt{3}$$) = 24 => h=12$$\sqrt{3}$$=(12*1.73)=20.76

Hence, the height of the tower is 20.76 m.

Ans .

18 m

1. Explanation :

Let AB be the tree and AC be the river. Let C and D be the two positions of the man. Then,

$$\angle$$ACB=$$60^{\circ}$$,$$\angle$$ADB=$$30^{\circ}$$ and CD=36 m.

Let AB=h metres and AC=x metres.

AB/AD=tan $$30^{\circ}$$=1/$$\sqrt{3}$$ => h/(36+x)=1/$$\sqrt{3}$$ =>h=(36+x)/$$\sqrt{3}$$

AB/AC=tan $$60^{\circ}$$=$$\sqrt{3}$$ => h/x=$$\sqrt{3}$$

h=$$\sqrt{3}$$/x .....(2)

From (i) and (ii), we get:

(36+x)/$$\sqrt{3}$$x => x=18 m.

So, the breadth of the river = 18 m.

Ans .

5 minutes

1. Explanation :

Let AB be the tower and C and D be the two positions of the boat.

h/y=tan $$60^{\circ}$$ =$$\sqrt{3}$$ => y=h/$$\sqrt{3}$$

h/(x+y)=tan $$30^{\circ}$$=1/$$\sqrt{3}$$ => x+y=$$\sqrt{3}$$h

x=(x+y)-y = ($$\sqrt{3}$$ h-h/$$\sqrt{3}$$ )=2h/$$\sqrt{3}$$

Now, 2h/$$\sqrt{3}$$ is covered in 10 min.

h/$$\sqrt{3}$$ will be covered in (10*($$\sqrt{3}$$/2h)*(h/$$\sqrt{3}$$))=5 min

Hence, required time = 5 minutes.

Ans .

18 m

1. Explanation :

Let AB and CD be the two temples and AC be the river.

Then, AB = 54 m. Let AC = x metres and CD=h metres.

$$\angle$$ACB=$$60^{\circ}$$, $$\angle$$EDB=$$30^{\circ}$$

AB/AC=tan $$60^{\circ}$$=$$\sqrt{3}$$

AC=AB/$$\sqrt{3}$$=54/$$\sqrt{3}$$=(54/$$\sqrt{3}$$*$$\sqrt{3}$$*$$\sqrt{3}$$)=18m

DE=AC=18$$\sqrt{3}$$

BE/DE=tan $$30^{\circ}$$=1/$$\sqrt{3}$$

BE=(18$$\sqrt{3}$$*1/$$\sqrt{3}$$=18 m

CD=AE=AB-BE=(54-18) m = 36 m.

So, Width of the river = AC = 18$$\sqrt{3}$$m=18*1.73 m=31.14m

Height of the other temple = CD= 18 m