CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 16

Answer: (d) 9 min

Explanation:
1. Let the cliff height be h. Distances: first \( h\cot 30° = h\sqrt{3} \), then \( h\cot 60° = \frac{h}{\sqrt{3}} \).
2. Distance covered in 6 min = \( h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \), so the speed is \( \frac{h}{3\sqrt{3}} \) per minute.
3. Total time = \( \frac{h\sqrt{3}}{h/(3\sqrt{3})} \) = 9 min (3 more minutes after the second observation).

Teacher's Note:
1. The angle of depression from the cliff equals the angle of elevation from the boat.
2. The remaining time after the second observation is 3 minutes; the total from the first is 9.

Answer: (d) 30°

Explanation:
1. At 60°, shadow s = \( \frac{h}{\sqrt{3}} \).
2. Longer shadow = 3s = \( h\sqrt{3} \), so tan θ = \( \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}} \) and θ = 30°.

Teacher's Note:
1. A longer shadow means a lower sun.
2. tan 30° = 1/√3.

Answer: (d) 56.05 m

Explanation:
1. Height of the first kite = \( 75\sin 45° = \frac{75}{\sqrt{2}} \) = 53.03 m.
2. Height of this kite above the roof = 53.03 − 25 = 28.03 m.
3. String = \( \frac{28.03}{\sin 30°} \) = 56.06 m, which is about 56.05 m.

Teacher's Note:
1. Height = string × sin(elevation).
2. The second boy's string is measured from the roof.

Answer: (d) 3 : 1

Explanation:
1. Let the middle point be M with MA = MB = x. Heights: \( x\tan 60° = x\sqrt{3} \) and \( x\tan 30° = \frac{x}{\sqrt{3}} \).
2. Ratio = \( \sqrt{3} : \frac{1}{\sqrt{3}} \) = 3 : 1.

Teacher's Note:
1. The horizontal distance to each tower is the same.
2. The ratio of heights equals the ratio of the tangents.

Answer: (c) 2

Explanation:
1. The distance from the centre to a corner is half the diagonal: \( \frac{a}{\sqrt{2}} \).
2. tan α = \( \frac{h\sqrt{2}}{a} \), so \( h\cot\alpha = \frac{a}{\sqrt{2}} \).
3. Hence \( \frac{a^2}{h^2\cot^2\alpha} = \frac{a^2}{a^2/2} \) = 2.

Teacher's Note:
1. Half the diagonal of a square of side a is a/√2.
2. cot α = (distance) ÷ h.

Answer: (a) \( 120\sqrt{6} \) m

Explanation:
1. Distance from A to the foot of the tower = 360 cot 60° = \( 120\sqrt{3} \) m. Distance from B to the foot = 360 cot 45° = 360 m.
2. AB is perpendicular to the bank, so the breadth w satisfies \( w^2 = 360^2 - (120\sqrt{3})^2 \) = 129600 − 43200 = 86400.
3. w = \( 120\sqrt{6} \) m.

Teacher's Note:
1. A and B are directly opposite each other, so triangle A-foot-B has a right angle at A.
2. Apply the Pythagoras theorem.

Answer: (a) \( \frac{h}{\sqrt{3}} \)

Explanation:
1. From the point h m above, the foot of the tower is seen at 60° depression, so the horizontal distance = \( h\cot 60° = \frac{h}{\sqrt{3}} \).

Teacher's Note:
1. Angle of depression from a height h gives distance = h cot(angle).
2. The 30° detail is not needed for this distance.

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

Explanation:
1. Let the cloud be H above the lake and d the horizontal distance. Then H − h = d tan α and H + h = d tan β.
2. Dividing: \( \frac{H + h}{H - h} = \frac{\tan\beta}{\tan\alpha} \), so \( H = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta - \tan\alpha} \).

Teacher's Note:
1. The image is as far below the water surface as the cloud is above it.
2. Use componendo and dividendo.

Answer: (c) \( \frac{r\sin\beta}{\sin(\alpha/2)} \)

Explanation:
1. The tangent from the eye makes angle α/2 with the line to the centre, so the distance eye-to-centre is \( \frac{r}{\sin(\alpha/2)} \).
2. Height = distance × sin β = \( \frac{r\sin\beta}{\sin(\alpha/2)} \).

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact.
2. The line to the centre bisects the angle subtended.

Answer: (c) 75°

Explanation:
1. Let the slope angle be θ and let the point be D m from the foot, beyond the base. Then tan 30° = \( \frac{c\sin\theta}{c\cos\theta + D} \) and tan 45° = \( \frac{2c\sin\theta}{2c\cos\theta + D} \).
2. Second: D = 2c(sin θ − cos θ). Put in the first: \( \sqrt{3}\sin\theta = \cos\theta + 2\sin\theta - 2\cos\theta \), so \( (2 - \sqrt{3})\tan\theta = 1 \).
3. tan θ = \( 2 + \sqrt{3} \) = tan 75°, so θ = 75°.

Teacher's Note:
1. The point and the foot of the slope lie on the same horizontal line.
2. tan 75° = 2 + √3.

Answer: (c) \( \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta}} \)

Explanation:
1. Let the height be H. The foot of the tower is at distance H cot α from A and H cot β from B, and these directions are perpendicular (south and east).
2. \( d^2 = H^2\cot^2\alpha + H^2\cot^2\beta \), so \( H = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta}} \).

Teacher's Note:
1. South and east are at right angles, so use the Pythagoras theorem.
2. Distance = H cot(elevation).

Answer: (d) \( \frac{n}{2n^2 + 1} \)

Explanation:
1. Let AB = H. Then tan∠APB = \( \frac{1}{n} \) and tan∠APC = \( \frac{1}{2n} \).
2. tan α = tan(∠APB − ∠APC) = \( \frac{\frac{1}{n} - \frac{1}{2n}}{1 + \frac{1}{2n^2}} = \frac{n}{2n^2 + 1} \).

Teacher's Note:
1. The angle subtended by CB is the difference of two elevations.
2. Use tan(A − B) = (tan A − tan B)/(1 + tan A tan B).

Answer: (d) \( \frac{h\tan p}{\tan q - \tan p} \)

Explanation:
1. Let d be the horizontal distance to the hill. Then d tan q − d tan p = h, so \( d = \frac{h}{\tan q - \tan p} \).
2. Height above the roof = d tan p = \( \frac{h\tan p}{\tan q - \tan p} \).

Teacher's Note:
1. The difference of the two heights seen from the two positions is the building height.
2. Here q is larger than p because the bottom is lower.

Answer: (c) \( \frac{\pi}{12} \)

Explanation:
1. Let the pole be h and t = tan α. Initial distance = h/t; new distance = \( \frac{h}{t} - 2h \) and tan 2α = \( \frac{h}{h/t - 2h} = \frac{t}{1 - 2t} \).
2. Using tan 2α = \( \frac{2t}{1 - t^2} \): 2(1 − 2t) = 1 − t², so t² − 4t + 1 = 0 and t = 2 − √3 = tan 15°.
3. α = 15° = π/12.

Teacher's Note:
1. tan 15° = 2 − √3.
2. Use the double-angle formula for tangent.

Answer: (b) \( x\tan(45° + \theta) \)

Explanation:
1. Using the cloud formula with β = 45°: \( H = \frac{x(\tan 45° + \tan\theta)}{\tan 45° - \tan\theta} = x\cdot\frac{1 + \tan\theta}{1 - \tan\theta} \).
2. This equals x tan(45° + θ).

Teacher's Note:
1. (1 + tan θ)/(1 − tan θ) = tan(45° + θ).
2. Reuse the cloud-and-reflection method.

Answer: (b) \( \frac{3a + b}{\sqrt{3}} \)

Explanation:
1. Radius of the first plot = a cot 30° = \( a\sqrt{3} \). Radius of the second = b cot 60° = \( \frac{b}{\sqrt{3}} \).
2. Distance between the centres = sum of the radii = \( a\sqrt{3} + \frac{b}{\sqrt{3}} = \frac{3a + b}{\sqrt{3}} \).

Teacher's Note:
1. The point of contact lies on the line of centres.
2. Externally touching circles: distance = r1 + r2.

Answer: (c) \( 250\sqrt{6} \) m

Explanation:
1. Let the height be H, so the first point is H from the foot (elevation 45°). After walking 500 m up the slope the position is 500 cos 15° horizontally and 500 sin 15° vertically.
2. tan 75° = \( \frac{H - 500\sin 15°}{H - 500\cos 15°} \). Solving gives H ≈ 612.4 m.
3. This equals \( 250\sqrt{6} \) m.

Teacher's Note:
1. Resolve the 500 m walk into horizontal and vertical parts.
2. 250√6 ≈ 612.4.

Answer: (c) \( 1 : \sqrt{3} \)

Explanation:
1. Let the pillar be h and the distance x, with t = tan θ = h/x for the pillar. The flagstaff adds angle θ, so the top of the flagstaff is at angle 2θ with height 3h: tan 2θ = 3h/x = 3t.
2. \( \frac{2t}{1 - t^2} = 3t \), so 2 = 3 − 3t², t² = 1/3 and t = \( \frac{1}{\sqrt{3}} \).
3. h : x = 1 : √3.

Teacher's Note:
1. Equal angles mean the top of the flagstaff is at twice the angle of the pillar top.
2. Cancel t from both sides.

Answer: (b) \( (150 - 20\sqrt{3}) \) m

Explanation:
1. Drop in height = 60 tan 30° = \( 20\sqrt{3} \) m.
2. Height of the first building = \( 150 - 20\sqrt{3} \) m.

Teacher's Note:
1. Angle of depression from the taller building's top is measured below the horizontal.
2. Subtract the drop from 150.

Answer: (a) \( 240\sqrt{3} \) km/h

Explanation:
1. Horizontal distances: first \( \cot 60° = \frac{1}{\sqrt{3}} \) km, then \( \cot 30° = \sqrt{3} \) km.
2. Distance covered = \( \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \) km in 10 s.
3. Speed = \( \frac{2}{\sqrt{3}} \times 360 = 240\sqrt{3} \) km/h.

Teacher's Note:
1. 10 s = 1/360 hour.
2. Take the difference of the two horizontal distances.

Answer: (a) 1.366 km

Explanation:
1. After the ascent the position is \( \frac{\sqrt{3}}{2} \) km horizontally and 0.5 km vertically from the foot. The summit is H high and H away horizontally from the foot.
2. tan 60° = \( \frac{H - 0.5}{H - \sqrt{3}/2} \) gives H − 0.5 = √3H − 1.5.
3. H = \( \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{2} \) = 1.366 km.

Teacher's Note:
1. Resolve the 1 km slope into horizontal and vertical parts.
2. Rationalise the denominator.

Answer: (b) \( \frac{a}{\cot\alpha + \cot\beta} \)

Explanation:
1. Let the height be h. The stations are h cot α and h cot β from the base, on opposite sides.
2. a = h(cot α + cot β), so h = \( \frac{a}{\cot\alpha + \cot\beta} \).

Teacher's Note:
1. The two distances add up when the stations are on opposite sides.
2. They would subtract if the stations were on the same side.

Answer: (b) \( 100(2 + \sqrt{3}) \) m

Explanation:
1. Using the formula, \( H = \frac{100(\tan 45° + \tan 30°)}{\tan 45° - \tan 30°} = 100\cdot\frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = 100\cdot\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \).
2. Rationalising, \( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{2} = 2 + \sqrt{3} \), so H = \( 100(2 + \sqrt{3}) \) m.

Teacher's Note:
1. The reflection lies as far below the surface as the object is above it.
2. Rationalise by multiplying by the conjugate.

Answer: (c) 20 m

Explanation:
1. Let the breadth be w and the tree height h. h = w tan 60° = \( w\sqrt{3} \) and h = (w + 40) tan 30° = \( \frac{w + 40}{\sqrt{3}} \).
2. 3w = w + 40, so w = 20 m.

Teacher's Note:
1. Equate the two expressions for the tree's height.
2. tan 60° = √3 and tan 30° = 1/√3.