Quantitative Aptitude
HEIGHT AND DISTANCE MCQs
Total Questions : 206
| Page 21 of 21 pages
Answer: Option C. -> 10 metres
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option A. -> 30°
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option C. -> 173 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.
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.
Answer: Option B. -> 60°
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°
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°
Answer: Option C. -> 273 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.
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.
Answer: Option 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.
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.
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