Arithmatical - Heights and Distance
DIRECTIONS : Problems based on Heights and Distance.
1. |
A ladder learning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall. |
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A. 9 m |
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B. 9.5 m |
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C. 10.5 m |
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D. 12 m |
Solution |
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Let AB be the wall and BC be the ladder. |
Then, < ABC | = 60° |
and, BC | = 19 m.; |
AC = x metres |
AC/BC | = cos 60° |
= x / 19 |
=1 / 2 |
x= 19/2 |
= 9.5 m. |
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2. |
The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is |
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A. 2.3 m |
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B. 4.6 m |
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C. 7.8 m |
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D. 9.2 m |
Solution |
Let AB be the wall and BC be the ladder. |
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Then, < ABC | = 60° |
AC | = 4.6 m.; |
AC/BC | = cos 60° |
= 1 /2 |
‹=›BC=2×AC |
=(2× 4.6) m |
= 9.2 m |
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3. |
The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree is |
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A. 30° |
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B. 45° |
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C. 60° |
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D. 90° |
Solution |
Let AB be the tree and AC be its shadow. |
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Then, < ABC | = θ. |
Then, AC/AB | = √3 |
cotθ= √3 |
θ=30° |
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4. |
If the height of a pole is 2√3 metres and the length of its shadow is 2 metres, find the angle of elevation of the sun. |
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A. 30° |
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B. 45° |
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C. 60° |
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D. 90° |
Solution |
Let AB be the pole and AC be its shadow. |
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Then, < ACB | = θ. |
Then, AB= 2 √3 m, AC = 2m, |
tanθ= AB/AC=2√3/2 |
= √3 |
θ=60° |
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5. |
From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is: |
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A. 149 m |
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B. 156 m |
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C. 173 m |
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D. 200 m |
Solution |
Let AB be the tower. |
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Then, <APB =30° and AB= 100 m |
AB/AP= tan30° |
= 1 / √3 |
AB = (AB x √3)= 100√3 m. |
= (100 x 1.73) m = 173 m. |
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