A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
A. 30°
B. 45°
C. 60°
D. 90°
Let AB be the tree and AC be its shadow.
Let ∠ACB = θ.
Then, AC/AB = √3
Cot θ = √3
θ = 30°
A. 149 m
B. 156 m
C. 173 m
D. 200 m
Let AB be the tower. Then, ∠APB = 30° and AB = 100 m, AB/AP = tan 30° = 1/√3 AP = (AB X √3)= 100√3 m. = (100 X 1.73) m = 173 m.
A. 50°
B. 60°
C. 70°
D. 80°
Let AB be the pole and AC be its shadow. Let angle of elevation, ∠ACB = θ. Then, AB = 2√3m, AC = 2 m. tan θ = AB/AC = 2√3/2 = √3 θ = 60°. So,the angle of elevation is 60°
A. 173 m
B. 200 m
C. 273 m
D. 300 m
Let AB be the lighthouse and C and D be the
positions of the ships. Then,
AB = 100 m, ∠ACB = 300 and ∠ADB = 45°.
AB/AC = tan 30° = 1/√3
AC = AB X √3 = 100√3 m.
AB/AD = tan 45° = 1 ⇒ AD = AB = 100 m.
CD = (AC + AD) = (100√3 + 100) m
= 100 (√3 +1) m = (100 X 2.73) m = 273 m.
A. 2.3 m
B. 4.6 m
C. 7.8 m
D. 9.2 m
Let AB be the wall and BC be the ladder. Then, ∠ACB = 60° and AC = 4.6 m. AC/BC = Cos 60° = 1/2 BC = 2 X AC = (2 X 4.6) m = 9.2 m.