Class 10 Heights and Distances — practice questions with answers
Trigonometry · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.
1. The angle made by the line of sight with the horizontal when we look UP at an object is called the angle of ____.
Aelevation
Bdepression
Cincidence
Dreflection
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Answer: A) elevation
When the object is above the horizontal level of the eye, the angle of the line of sight above the horizontal is the angle of elevation.
2. From the top of a 50 m tower, the angle of depression of a car on the road is 30 deg. The distance (in m) of the car from the foot of the tower is:
A50 3
B350
C25 3
D100
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Answer: A) 50 3
Depression 30 = elevation 30 at car. tan 30 = d50 => 31 = d50 => d = 50 3 m.
3. In the previous setup, a man on a 10 m deck sees a lighthouse at horizontal distance 10 3 m. The top is at elevation 60 deg above deck level. Find the total height of the lighthouse above sea level (in m).
A80
B40
C44
D36
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Answer: B) 40
Above deck: tan 60 = 103x => x = 10 3 x 3 = 30. Add deck height 10 => total = 30 + 10 = 40 m.
4. The angle of elevation of the top of a tower from a point A on the ground is 30 deg. On walking 30 m toward the tower to point B, the angle of elevation becomes 60 deg. Find the height of the tower (in m). (Take sqrt(3) = 1.732.)
A22.98
B51.96
C25.98
D28.98
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Answer: C) 25.98
Let height h, BB' = x. tan60 = xh => x = 3h. tan30 = x+30h => x+30 = h 3. So h 3 - 3h = 30 => 32h = 30 => h = 15 3 = 15 x 1.732 = 25.98 m.
5. At 8 a.m. in Chennai the angle of elevation of the Sun is 30 deg, and by noon it has become 60 deg. A boy 1.5 m tall stands on the same level ground at both times. By how many metres does his shadow shorten between 8 a.m. and noon? Give the exact value.
A1.5 3
B3
C0.5 3
D3
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Answer: B) 3
At 30 deg the shadow is tan301.5 = 1.5 3 m (about 2.60 m). At 60 deg it is 31.5 = 0.5 3 m (about 0.87 m). Shortening = 1.5 3 - 0.5 3 = 3 m (about 1.73 m).