KnowledgeBoat Logo
|

Mathematics

Directions:

The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. The tower is 30 metres high.

Based on this information, answer the following questions:

34.The horizontal distance between the tower and the building is:
(a) 10 m
(b) 17.3 m
(c) 20 m
(d) 34.6 m

35.The height of the building is:
(a) 10 m
(b) 10310\sqrt{3} m
(c) 15 m
(d) 15315\sqrt{3} m

36. The straight line distance between the tops of the tower and the building is:
(a) 10310\sqrt{3} m
(b) 10510\sqrt{5} m
(c) 10710\sqrt{7} m
(d) 30 m

37. A bird flew straight from the top of the tower to the foot of the building. What is the distance that the bird flew?
(a) 20 m
(b) 25 m
(c) 20320\sqrt{3} m
(d) 20(31)20(\sqrt{3}-1) m

Heights & Distances

3 Likes

Answer

A bird flew straight from the top of the tower to the foot of the building. What is the distance that the bird flew? Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

34. Let CD be the height of the tower and h be the height of the building (AB).

Let x (BD) be the horizontal distance between them.

In triangle CDB,

tan60=30x3=30xx=303x=3033x=103=17.3 m.\Rightarrow \tan 60^{\circ} = \dfrac{30}{x} \\[1em] \Rightarrow \sqrt3 = \dfrac{30}{x} \\[1em] \Rightarrow x = \dfrac{30}{\sqrt3} \\[1em] \Rightarrow x = \dfrac{30\sqrt3}{3} \\[1em]\Rightarrow x = 10 \sqrt3 = 17.3 \text{ m}.

Hence, option (b) is the correct option.

35.In triangle ADB,

tan30=hx13=h103h=10 m.\Rightarrow \tan 30^{\circ} = \dfrac{h}{x} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{10\sqrt3} \\[1em] \Rightarrow h = 10 \text{ m}.

Hence, option (a) is the correct option.

36. In right-angled triangle △AEC at E.

AE = BD = 10310\sqrt3 m

CE = CD - AB = 30 - 10 = 20 m

∴ AC2 = AE2 + CE2

AC2 = (10310\sqrt3)2 + (20)2

AC2 = 100(3) + 400

AC2 = 300 + 400

AC2 = 700

AC = 700=107\sqrt{700} = 10 \sqrt7 m

Hence, option (c) is the correct option.

37. In triangle CDB,

sin60=CDBC32=30BCBC=603BC=6033BC=203 m.\Rightarrow \sin 60^{\circ} = \dfrac{CD}{BC} \\[1em] \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{30}{BC} \\[1em] \Rightarrow BC = \dfrac{60}{\sqrt3} \\[1em] \Rightarrow BC = \dfrac{60\sqrt3}{3} \\[1em] \Rightarrow BC = 20\sqrt3 \text{ m.}

Hence, option (c) is the correct option.

Answered By

1 Like


Related Questions