Directions :

A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

Based on this information, answer the following questions:

38.The height of the tower is:
(a) 10 m
(b) $10\sqrt{3}$ m
(c) 20 m
(d) $20\sqrt{3}$ m

39.The width of the canal is:
(a) 10 m
(b) $10\sqrt{3}$ m
(c) 20 m
(d) $20\sqrt{3}$ m

40.The straight line distances of the top of the tower from the two points of observation differ by:
(a) 7.32 m
(b) 14.64 m
(c) 17.32 m
(d) 20.64 m

41.How far away from the other bank must the point of observation be, so that the angle of elevation of the top of the tower is 45°? (a) 7.32 m
(b) 10 m
(c) 14.64 m
(d) 27.32 m

38. Let hbe the height of the tower (AB) and w be width of canal (BC).

Let D be another point 20 m away from C.

In triangle ABC,

$\Rightarrow \tan 60^{\circ} = \dfrac{h}{w} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{w} \\[1em] \Rightarrow w = \dfrac{h}{\sqrt3} …..(1)$

In triangle ADB,

$\Rightarrow \tan 30^{\circ} = \dfrac{h}{w + 20} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{w + 20} \\[1em] \Rightarrow w + 20 = h\sqrt3 \\[1em] \Rightarrow \dfrac{h}{\sqrt3} + 20 = h\sqrt3 \\[1em] \Rightarrow \dfrac{h + 20\sqrt{3}}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow h + 20\sqrt{3} = 3h \\[1em] \Rightarrow 20\sqrt{3} = 2h \\[1em] \Rightarrow h = 10\sqrt{3} \text{ m}.$

Hence, option (b) is the correct option.

39. In triangle ABC,

$\Rightarrow \tan 60^{\circ} = \dfrac{h}{w} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{w} \\[1em] \Rightarrow w = \dfrac{h}{\sqrt3} \\[1em] \Rightarrow w = \dfrac{10\sqrt3}{\sqrt3} \\[1em] \Rightarrow w = 10 \text{ m.}$

Hence, option (a) is the correct option.

40. In triangle ABC,

$\Rightarrow \sin 60^{\circ} = \dfrac{h}{AC} \\[1em] \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{10\sqrt3}{AC} \\[1em] \Rightarrow AC = 20 \text{ m.}$

In triangle ABD,

$\Rightarrow \sin 30^{\circ} = \dfrac{h}{AD} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{10\sqrt3}{AD} \\[1em] \Rightarrow AD = 20\sqrt3 \text{ m.}$

The straight line distances of the top of the tower from the two points of observation differ by = AD - AC = $20\sqrt{3} - 20$

= $20(\sqrt{3} - 1)$

= 20(0.732) = 14.64 m.

Hence, option (b) is the correct option.

41. Let E be the new point of observation

In triangle ABE,

$\Rightarrow \tan(45^\circ) = \dfrac{h}{BE} \\[1em] \Rightarrow 1 = \dfrac{10\sqrt{3}}{BE} \\[1em] \Rightarrow BE = 10\sqrt{3} \text{ m}$

Distance from the bank (CE) = BE - BC

CE = $10\sqrt{3}$ - 10

CE = 10($\sqrt{3}$ - 1)

CE = 10(0.732) = 7.32 m

Hence, option (a) is the correct option.