CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 07

Answer: (a) \( 20(\sqrt{3}-1) \) m

Explanation:
1. Let the point be O, the foot of the building A, and the tower height h. From the bottom of the tower, tan 45° = 20/OA, so OA = 20 m.
2. From the top of the tower, \( \tan 60^\circ = \frac{20+h}{OA} = \frac{20+h}{20} \), so \( \sqrt{3} = \frac{20+h}{20} \).
3. Hence h = \( 20(\sqrt{3}-1) \) m.

Teacher's Note:
1. First find the horizontal distance from the 45° angle.
2. The 60° angle uses the combined height of building and tower.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. In the right triangle ACD, \( \tan\theta = \frac{CD}{AC} \).
2. In the right triangle ACB, \( \tan\phi = \frac{BC}{AC} \).
3. Dividing, \( \frac{\tan\theta}{\tan\phi} = \frac{CD}{BC} = \frac{1}{2} \), because D is the mid-point of BC. The answer is \( \frac{1}{2} \).

Teacher's Note:
1. Write both tangents as ratios of the same vertical side AC.
2. The common AC cancels, leaving CD : BC.

Answer: (b) \( \sqrt{xy} \)

Explanation:
1. Let the height be h and the angle at A be θ, so the angle at B is 90° − θ.
2. Then \( \tan\theta = \frac{h}{x} \) and \( \tan(90^\circ-\theta) = \cot\theta = \frac{h}{y} \).
3. Multiplying, \( \tan\theta\cot\theta = \frac{h^2}{xy} \), so \( h^2 = xy \) and h = \( \sqrt{xy} \).

Teacher's Note:
1. For complementary angles, tan θ × cot θ = 1.
2. The height is the geometric mean of the two distances.

Answer: (d) 6.96 m

Explanation:
1. Let the broken part be x m, so the standing part is (15 − x) m.
2. The broken part is the hypotenuse of a right triangle, and sin 60° = \( \frac{15-x}{x} \), so \( \frac{\sqrt{3}}{2} = \frac{15-x}{x} \).
3. Solving, \( x = \frac{30}{2+\sqrt{3}} = 30(2-\sqrt{3}) \approx 8.04 \) m, so the height is 15 − 8.04 ≈ 6.96 m.

Teacher's Note:
1. Treat the standing part as the side opposite the 60° angle.
2. Rationalise 30/(2 + √3) by multiplying by (2 − √3).

Answer: (a) 75 m

Explanation:
1. At the same time the sun's elevation is the same, so the triangles are similar.
2. Hence \( \frac{6}{4} = \frac{h}{50} \).
3. So h = \( \frac{6\times 50}{4} \) = 75 m.

Teacher's Note:
1. Height divided by shadow is the same for all objects at one time of day.
2. That ratio equals tan of the sun's elevation.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. Let the foot of the tower be B. The first point is at distance \( \cot\alpha \cdot h \) from B, because \( \cot\alpha = \frac{\text{distance}}{h} \).
2. Similarly the second point is at distance \( h\cot\beta \) from B.
3. The difference of the distances is h/2, so \( h\cot\alpha - h\cot\beta = \frac{h}{2} \) and \( \cot\alpha-\cot\beta = \frac{1}{2} \).

Teacher's Note:
1. Cotangent is base divided by height.
2. The distance moved equals the difference of the two base lengths.

Answer: (a) \( 75\sqrt{2} \) m

Explanation:
1. The width x is the side opposite the 45° angle, and the bridge is the hypotenuse.
2. \( \sin 45^\circ = \frac{x}{150} \Rightarrow \frac{1}{\sqrt{2}} = \frac{x}{150} \).
3. So x = \( \frac{150}{\sqrt{2}} = 75\sqrt{2} \) m, which is \( 75\sqrt{2} \) m.

Teacher's Note:
1. The width of the river is perpendicular to the banks.
2. Use sine when the hypotenuse is given.

Answer: (b) \( \sqrt{2}-1 \)

Explanation:
1. Let the height be h and let the final position be x from the tower, so \( \tan(90^\circ-\alpha) = \cot\alpha = \frac{h}{x} \) and x = h tan α.
2. From the first position, \( \tan\alpha = \frac{h}{2h+x} = \frac{h}{2h+h\tan\alpha} \).
3. This gives \( \tan^2\alpha + 2\tan\alpha - 1 = 0 \), so \( \tan\alpha = -1+\sqrt{2} \) (the positive root), which is \( \sqrt{2}-1 \).

Teacher's Note:
1. Write both distances in terms of h and tan α.
2. Reject the negative root because tan α is positive for an acute angle.

Answer: (a) 4.28 m, 2.14 m

Explanation:
1. The height to be reached is 5 − 1.3 = 3.7 m.
2. The ladder is the hypotenuse, so \( \sin 60^\circ = \frac{3.7}{L} \) and \( L = \frac{3.7\times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28 \) m.
3. The distance of the foot from the pole is \( \frac{3.7}{\tan 60^\circ} = \frac{3.7}{1.73} \approx 2.14 \) m. So the answer is 4.28 m, 2.14 m.

Teacher's Note:
1. Subtract 1.3 m from the pole height first.
2. Use sine for the ladder length and tangent for the distance on the ground.

Answer: (d) 3 cm

Explanation:
1. In the right triangle ABC, the hypotenuse AC = 6 cm and BC is opposite the 30° angle.
2. \( \sin 30^\circ = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{BC}{6} \).
3. So BC = 3 cm.

Teacher's Note:
1. The side opposite 30° is half of the hypotenuse.
2. Use sine when the hypotenuse is given.

Answer: (c) \( 75\sqrt{3} \) m

Explanation:
1. The angle of elevation of the top from the object is also 30°, so \( \tan 30^\circ = \frac{75}{d} \).
2. \( \frac{1}{\sqrt{3}} = \frac{75}{d} \), so d = \( 75\sqrt{3} \) m.

Teacher's Note:
1. The angle of depression equals the angle of elevation from the object (alternate angles).
2. Use tan = opposite / adjacent.

Answer: (a) 45°

Explanation:
1. Let the height and the shadow both be x, and the angle of elevation of the sun be θ.
2. Then tan θ = x/x = 1, so θ = 45°.

Teacher's Note:
1. Equal height and shadow always mean tan θ = 1.
2. tan 45° = 1.

Answer: (d) 15 m

Explanation:
1. Let the height be h. Then tan 45° = h/15.
2. Since tan 45° = 1, h = 15 m.

Teacher's Note:
1. At 45° the height equals the shadow length.
2. Use tan = height / shadow.

Answer: (c) 60°

Explanation:
1. The angle of depression is measured from the horizontal line through the observer down to the line of sight.
2. The horizontal through the observer is parallel to the ground, so the angle of depression equals the angle of elevation of the observer from the object.
3. In the right triangle the angles are 30°, 60° and 90°, and the angle at the object is 60°. So the angle of depression is 60°.

Teacher's Note:
1. Alternate angles between parallel lines are equal.
2. Angle of depression = angle of elevation from the object.

Answer: (b) \( \frac{h(\tan\alpha+\tan\beta)}{\tan\beta} \)

Explanation:
1. Let the distance between the building and the tower be x. From the depression, \( \tan\beta = \frac{h}{x} \), so x = \( \frac{h}{\tan\beta} \).
2. Above the level of the building top, the tower rises y where \( \tan\alpha = \frac{y}{x} \), so y = \( \frac{h\tan\alpha}{\tan\beta} \).
3. Height of tower = h + y = \( h\left(1+\frac{\tan\alpha}{\tan\beta}\right) = \frac{h(\tan\alpha+\tan\beta)}{\tan\beta} \).

Teacher's Note:
1. The tower height is the building height plus the part above the building top.
2. Both parts use the same horizontal distance x.

Answer: (d) θ₁ and θ₂ are always equal.

Explanation:
1. The line OA is above the horizontal line OC, so θ₁ is the angle of elevation. The statement (a) is correct.
2. The line OB is below the horizontal, so θ₂ is the angle of depression. The statement (b) is correct, and so is (c) by definition.
3. The two angles depend on where the objects A and B are, so they are not always equal. The incorrect statement is (d).

Teacher's Note:
1. Elevation is above the horizontal; depression is below it.
2. Both are measured from the horizontal line.

Answer: (d) None of these

Explanation:
1. Statement (a) is the definition of the line of sight, so it is correct.
2. Statements (b) and (c) are the definitions of the angle of elevation and the angle of depression, so they are correct.
3. All the given statements are correct, so none of them is incorrect: option (d).

Teacher's Note:
1. Learn the three definitions word for word.
2. Elevation is for an object above the eye level, depression for one below it.

Answer: (b) 45°

Explanation:
1. tan θ = length of rod / length of shadow = 1 / 1 = 1.
2. So θ = 45°.

Teacher's Note:
1. Equal rod and shadow lengths give tan θ = 1.
2. Remember tan 45° = 1.

Answer: (c) angle of depression

Explanation:
1. When the object is below the horizontal level, the observer lowers the head to look at it.
2. The angle between the line of sight and the horizontal is then the angle of depression.

Teacher's Note:
1. Object above the eye level gives the angle of elevation.
2. Object below the eye level gives the angle of depression.