Subjective Questions
Q1:
A man observes the top of the tower and finds the angle of elevation to be 30°. If the tower is 30m away from him, find the height of the tower?
Type: Short
Difficulty: Easy
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Answer: <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe88.png" alt="." width="200" height="173"><figcaption><br></figcaption></figure><p>Solution:<br>Let MN be the height of the tower and NO be the distance between the tower and man respectively.<br>Here,<br>O is the point of observation.<br>∠MON =30°, NO = 30m and MN = ?</p> <p>From the right angled triangle MNO,<br>or, tan30° = \(\frac{MN}{NO}\)<br>or, \(\frac{1}{\sqrt{3}}\) = \(\frac{MN}{30}\)<br>or, MN = \(\frac{30}{\sqrt{3}}\)<br>∴ The height of the tower is\(\frac{30}{\sqrt{3}}\)m.</p>
Q2:
The height of a building is 15m. The angle of elevation of the top of a building as seen from a point on the level of a ground is 60°. Find the distance of the point from the foot of the building.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/zxc44.png" alt="/" width="200" height="173"><figcaption><br></figcaption></figure><figure class="inline-right" style="width: 200px;"><br><figcaption><br></figcaption></figure><p>Let EF be a height of a building and FG be the distance between the point on the ground level and foot of the building.<br>Here,<br>∠EFG = 60°, EF = 15m and EG = ?</p> <p>From the right angled triangle EFG<br>or, tan60° = \(\frac{EF}{FG}\)<br>or, \(\sqrt{3}\) = \(\frac{15}{FG}\)<br>or, FG = \(\frac{15}{\sqrt{3}}\)<br>∴ The required distance is \(\frac{15}{\sqrt{3}}\)m.</p>
Q3:
A vertical pole is 10ft high and the length of its shadow is 10\(\sqrt{3}\) ft. What is the altitude of the sun?
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe90.png" alt="." width="200" height="173"><figcaption><br></figcaption></figure><p>Let XY be the height of a pole and ZY be the length of its shadow. Let ∠XYZ =α be the altitude of the sun.<br>Here,<br>XY = 10ft, ZY = 10\(\sqrt{3}\) ft, α = ?</p> <p>From the right angled triangle XYZ,<br>or, tanα = \(\frac{XY}{ZY}\)<br>or, tanα = \(\frac{10}{10\sqrt{3}}\)<br>or, tanα = \(\frac{1}{\sqrt{3}}\)<br>or, tanα = tan30°<br>∴ α = 30°<br>Hence, the altitude of te sun is 30°.</p> <p><strong>Note: </strong>altitude of the sun = angle of elevation of the sum.</p>
Q4:
From the top of a tree 20m high, a man observes the angle of depression of an object and finds it to be 45°. Find the distance between the tree and the object?
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe91.png" alt="." width="200" height="173"><figcaption><br></figcaption></figure><p>Let OP be the height of the tree and QP be the distance between the distance the object and the tree.<br>Here,<br>O is the point of observation.<br>∠AOQ = ∠OQP =45°, OP = 20m and QP = ?</p> <p>From the right angled triangle OQP,<br>or, tan45° = \(\frac{OP}{QP}\)<br>or, 1 = \(\frac{20}{QP}\)<br>or, QP = 20<br>Hence, the distance between the object and the tree is 20m.</p>
Q5:
A man 1.6m tall observes the top of tower 61.1m high situated in front of him and finds an angle of elevation on 60°. How far is a man from the tower?
Type: Long
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe92.png" alt="." width="200" height="150"><figcaption><br></figcaption></figure><p>Height of tower =AB - BD = (61.6 − 1.6)m = 60m<br>Now,<br>or, tan60° = \(\frac{p}{b}\)<br>or, \(\sqrt{3}\) = \(\frac{60}{BC}\)<br>or, \(\sqrt{3}\)BC = 60<br>or, BC = \(\frac{60}{\sqrt{3}}\)<br>or,BC = \(\frac{60}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both sides by\(\sqrt{3}\) )<br>or, BC = \(\frac{60\sqrt{3}}{3}\)<br>or, BC = 20\(\sqrt{3}\)<br>∴The distance between man and tower is 20\(\sqrt{3}\)m</p>
Q6:
A man 1.7m tall is standing 20m away from a pole on the same level of the ground. He observes the angle of elevation of the top of the pole and finds to be 30°. Calculate the height of the pole.
Type: Long
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe93.png" alt="." width="200" height="150"><figcaption><br></figcaption></figure><p>Let, height of a pole be AD, angle of elevation is 30°, height of man is 1.7m<br>Given,<br>∠C = 30°, CE = BD = 1.7m and CB = ED = 20m</p> <p>From right angle triangle ABC<br>or, tan30° = \(\frac{AB}{BC}\)<br>or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AB}{20}\)<br>or, \(\sqrt{3}\) AB = 20<br>or, AB = \(\frac{20}{\sqrt{3}}\)<br>or, AB = \(\frac{20}{\sqrt{3}}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both side by \(\sqrt{3}\)<br>or, AB = \(\frac{20\sqrt{3}}{3}\)<br>∴ AB = 11.54m<br>Now,<br>AD = AB + BD<br> = 11.54 + 1.7 m<br> = 13.24m<br>∴ Height of pole is 13.24m.</p>
Q7:
A tree of 14m height is broken by the wind so that its top touches the ground and makes an angle of 60° with the ground. Find the length of the broken part of the tree.
Type: Long
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe94.png" alt="." width="200" height="150"><figcaption><br></figcaption></figure><p>Let lenght of the tree be CA<br>From right angle triangle<br>or, sinθ = \(\frac{p}{h}\)<br>or, sin60° = \(\frac{14 − x}{x}\)<br>or, \(\frac{\sqrt{3}}{2}\) = \(\frac{14 − x}{x}\)<br>or, \(\sqrt{3}\) x = 28 − 2x<br>or, 1.73x = 28 − 2x<br>or, (1.73 + 2) x = 28<br>or, 3.37x = 28<br>or, x = \(\frac{28}{3.37}\)<br>∴ x = 7.5m<br>∴ The lenght of a tree is 7.5m.</p>
Q8:
A ladder 200ft long rest against a vertical wall. If the foot of the ladder is 100\(\sqrt{3}\) ft from the base of the wall, find the angle made by the ladder with the horizontal ground.
Type: Long
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe95.png" alt="." width="200" height="148"><figcaption><br></figcaption></figure><p>Let, AB be the angle made by the ladder with the horizontal ground and AC is a ladder of 200ft and BC is the base of wall.<br>Here,<br>AC = 200ft, CB = 100\(\sqrt{3}\) ft and AB = ?<br>By using Pythagoros Theorem<br>or, AB = \(\sqrt{AC^2− CB^2}\)<br>or, AB = \(\sqrt{200^2− (100\sqrt{3})^2}\)<br>or, AB = \(\sqrt{40000− 30000}\)<br>or, AB = \(\sqrt{10000}\)<br>or, AB = 100ft<br>Again,<br>or, sinC = \(\frac{AB}{AC}\)<br>or, sinC = \(\frac{100}{200}\)<br>or, sinC = sin 30°<br>∴ C = 30°<br>Hence, the angle made by the ladder with the horizontal ground is 30°.</p>
Q9:
The angle of elevation of the top a pole at distance 30m is 60°, find the height of the tree.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/zxc45.png" alt="." width="200" height="166"><figcaption><br></figcaption></figure><figure class="inline-right" style="width: 200px;"><br><figcaption><br></figcaption></figure><p>Let, BC be the distance between tree, AC be height of tower and point is 30m and angle of elevation is 60°<br>Given,<br>∠B = 60° and BC = 30m<br>Now,<br>or, tan60° = \(\frac{AC}{BC}\)<br>or, \(\sqrt{3}\) = \(\frac{AC}{30}\)<br>∴ AC = 30\(\sqrt{3}\)m<br>The height of tree is 30\(\sqrt{3}\)m.</p>
Q10:
The height of a tree is 15m. The angle of elevation of the top of the tree as seen from a point on the ground level is 45°. Find the distance of the point from the foot of the tree.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/zxc46.png" alt="/" width="200" height="149"><figcaption><br></figcaption></figure><figure class="inline-right" style="width: 200px;"><br><figcaption><br></figcaption></figure><p>Let, YZ be the distance between the point to the foot of a tree and XY is the height of tree which is 15m and the angle of elevation is 45°<br>Here,<br>XY = 15m, ∠XZY = 45° and YZ = ?<br>Now,<br>In right angle triangle XYZ,<br>or, tan45° = \(\frac{XY}{YZ}\)<br>or, 1 = \(\frac{15}{YZ}\)<br>or, YZ = 15m<br>∴ The distance of the point from the foot of the tree is 15m.</p>
Q11:
The upper part of a straight tree is broken by the wind and makes an angle of 45° with the plain surface at a point 9m from the foot of the tree. Find the height of the tree before it was broken.
Type: Long
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qwe99.png" alt="." width="200" height="161"><figcaption><br></figcaption></figure><p>Let, height of tree be AB<br>From right angle triangle<br>or, tan45° = \(\frac{p}{b}\)<br>or, 1 = \(\frac{AB}{9}\)<br>∴ AB = 9m<br>Again,<br>or, cos45° = \(\frac{b}{h}\)<br>or, \(\frac{1}{\sqrt{2}}\) = \(\frac{9}{BC}\)<br>∴ BC = 9\(\sqrt{2}\)<br>Then,<br>AC = AB + CB<br>= 9 + 9\(\sqrt{2}\)<br>= 9 + 12.72<br>= 21.72m<br>Hence, the height of the tree before it was broken is 21.72m</p>
Q12:
The string of a kite is inclined to the ground at an angle of 30°. The length of the string is 15m. Find the height of the kite.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qaz2.png" alt="." width="209" height="170"><figcaption><br></figcaption></figure><p>Let, RQ be the point between the point of the foot of a tree and PR be the height of the tree which is 15m and angle of elevation is 30°<br>Here,<br>PR = 15m, ∠PQR = 30° and QR = ?<br>Now,<br>In right angled triangle<br>or, tan30° = \(\frac{PR}{QR}\)<br>or, \(\frac{1}{\sqrt{3}}\) = \(\frac{15}{QR}\)<br>or QR = 15\(\sqrt{3}\)<br>∴ The height of a kite is15\(\sqrt{3}\)m.</p>
Q13:
The angle of elevation of a top of a tower at a distance of a 100m is found to be 45°. Find the height of the tower.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qaz3.png" alt="." width="200" height="161"><figcaption><br></figcaption></figure><p>Let, AB be the height of a tower and BC the distance between the tower and a point is 100m and angle of elevation is 45°<br>Here,<br>BC = 100m, ∠ABC = 45° and AB = ?<br>In right angle triangle<br>or, tan45° = \(\frac{p}{b}\)<br>or, 1 = \(\frac{AC}{100}\)<br>or, AC = 100m<br>∴ The height of the tower is 100m.</p>
Q14:
What is the altitude of the sun, when the shadow of a pillar of height 30ft is 30\(\sqrt{3}\) ft?
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qaz4.png" alt="." width="200" height="161"><figcaption><br></figcaption></figure><p>Let, AB be the height of a pillar and AC is the shadow of pillar of height 30ft and angle of elevation is θ<br>Here,<br>AB = 30ft, BC = 30\(\sqrt{3}\) and ∠C = ?<br>Now,<br>or, tanθ = \(\frac{AB}{BC}\)<br>or, tanθ = \(\frac{30}{30\sqrt{3}}\)<br>or, tanθ = \(\frac{1}{\sqrt{3}}\)<br>or, tanθ = tan30°<br>∴ θ =30°<br>Hence,the altitude of the sun is 30°.</p>
Q15:
The angle of elevation of top of a pole at distance 200\(\sqrt{3}\)m from its foot is found to be 30°. Find the height of a pole.
Type: Short
Difficulty: Easy
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Answer: <p>Solution:</p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/qaz6.png" alt="." width="200" height="161"><figcaption><br></figcaption></figure><p></p> <p>Let AC be height of pole, BC be distance between pole of point is200\(\sqrt{3}\)m and angle of elevation is 30°<br>Given,<br>∠B = 30°, BC = 200\(\sqrt{3}\)m and AC = ?<br>Now,<br>or, tan30° = \(\frac{AC}{BC}\)<br>or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AC}{200\sqrt{3}}\)<br>or, AC =\(\frac{200\sqrt{3}}{\sqrt{3}}\)<br>or, AC = 200<br>∴ The height of the pole is 200m.</p>
