CBSE Class 10 Mathematics Heights and Distances Worksheet Set 07

Answer: tan θ = 3/√3 = √3, so θ = 60°.

Teacher's Note:
1. Height and shadow are the two perpendicular sides.
2. tan 60° = √3.

Answer: tan θ = height/shadow = 1, so θ = 45°.

Teacher's Note:
1. The ratio of height to shadow is the tangent of the elevation.
2. tan 45° = 1.

Answer: Height of tower above the eye = 22 − 1.5 = 20.5 m, and the horizontal distance is 20.5 m. So tan θ = 1 and θ = 45°.

Teacher's Note:
1. Subtract the observer's height from the tower height.
2. Equal opposite and adjacent sides give 45°.

Answer: Height = 15 sin 60° = 15 × √3/2 = 7.5√3 m ≈ 12.99 m.

Teacher's Note:
1. The wall height is opposite to the angle at the ground.
2. Use sine because the ladder is the hypotenuse.

Answer: BC = AB tan 60° = 20√3 cm.

Teacher's Note:
1. In right triangle ABC, BC is opposite to angle A.
2. Use tan 60° = √3.

Answer: Height = 20 tan 30° = \( \dfrac{20}{\sqrt{3}} \) m ≈ 11.55 m.

Teacher's Note:
1. Height is opposite and the distance is adjacent.
2. tan 30° = 1/√3.

Answer: tan θ = h/(√3 h) = 1/√3, so θ = 30°.

Teacher's Note:
1. Divide height by shadow length.
2. tan 30° = 1/√3.

Answer: The elevation θ has tan θ = 10/(10√3) = 1/√3, so θ = 30°. Tower height = 90 tan 30° = 30√3 m ≈ 51.96 m.

Teacher's Note:
1. The sun's angle is the same for both objects.
2. Use it to find the tower's height.

Answer: Height = 50 sin 60° = 25√3 m ≈ 43.3 m.

Teacher's Note:
1. Use sine with the ladder as hypotenuse.
2. sin 60° = √3/2.

Answer: The difference in heights is 5 m and the horizontal gap is 12 m. Distance = √(25 + 144) = 13 m.

Teacher's Note:
1. Form a right triangle with the height difference and ground distance.
2. Apply Pythagoras theorem.

Answer: Let height = h. Shadows are h and h/√3, so h − h/√3 = 10, giving h = 10√3/(√3 − 1) = 5√3(√3 + 1) = 5(3 + √3) m ≈ 23.66 m.

Teacher's Note:
1. Write each shadow length using cot or tan.
2. Rationalise the denominator to simplify.

Answer: Horizontal distance d = 100/tan 60° = 100/√3. The tower's top is below the cliff top by d tan 30° = 100/3. Height of tower = 100 − 100/3 = \( \dfrac{200}{3} \) m ≈ 66.67 m.

Teacher's Note:
1. Find the horizontal distance using the bottom angle.
2. Use the top angle to find the drop to the tower top.

Answer: Width = 9/tan 60° = 3√3 m. Height above the window = 3√3 tan 30° = 3 m. Height of the house = 9 + 3 = 12 m, width = 3√3 m ≈ 5.2 m.

Teacher's Note:
1. The foot angle gives the street width from the window height.
2. Add the part above the window to 9 m.

Answer: With height h km: h cot 30° − h cot 45° = 1, so h(√3 − 1) = 1 and h = (√3 + 1)/2 ≈ 1.37 km.

Teacher's Note:
1. The nearer stone has the larger angle of depression.
2. The difference of the two horizontal distances is 1 km.

Answer: Let the point be x m from the first pole. h = x tan 60° = (80 − x) tan 30°, so 3x = 80 − x and x = 20. Height = 20√3 m; distances are 20 m and 60 m.

Teacher's Note:
1. Use the same height h in both right triangles.
2. Solve the equation in x.

Answer: Speed = 200 m/s, so distance covered = 3000 m. With height h: h cot 30° − h cot 60° = 3000, i.e. h(√3 − 1/√3) = 3000, so h = 1500√3 m ≈ 2598 m.

Teacher's Note:
1. Convert km/h to m/s first.
2. Equate the difference of horizontal distances to the distance flown.

Answer: PB = 10/tan 30° = 10√3 m. Height = 10√3 tan 60° = 30 m.

Teacher's Note:
1. Find the distance from P to the building using the window.
2. Use the same distance for the larger triangle.

Answer: Distance = 14/tan 30° = 14√3 m. Height above the deck = 14√3 tan 60° = 42 m. Height of cliff = 42 + 14 = 56 m.

Teacher's Note:
1. The base angle gives the horizontal distance.
2. Add the deck height to the part above the deck.

Answer: Shadows: 50√3 m at 30° and 50 m at 45°. x = 50(√3 − 1) = 36.6 m.

Teacher's Note:
1. Use cot to find each shadow length.
2. Subtract the two.

Answer: Horizontal distance = 3125/tan 30° = 3125√3. Height of upper plane = 3125√3 × √3 = 9375 m. Gap = 9375 − 3125 = 6250 m.

Teacher's Note:
1. The lower plane has the smaller angle of elevation.
2. Both planes are above the same ground point.

Answer: Distance = 60/tan 60° = 20√3 m. Drop to the post top = 20√3 tan 30° = 20 m, so height of post = 60 − 20 = 40 m.

Teacher's Note:
1. The bottom angle gives the horizontal distance.
2. Subtract the drop to the post top from 60 m.

Answer: Let tower height = H and distance = d. tan α = H/d and tan β = (H + h)/d. Then d = H/tan α, so tan β = (H + h) tan α/H, giving H(tan β − tan α) = h tan α and H = h tan α/(tan β − tan α).

Teacher's Note:
1. Write two equations using the same horizontal distance.
2. Eliminate d and solve for H.

Answer: Let the cloud be H m above the lake. Height above the observer = H − 60; the reflection is H + 60 below him. Horizontal distance d = (H − 60)√3 = (H + 60)/√3, so 3(H − 60) = H + 60 and H = 120 m.

Teacher's Note:
1. The reflection is as far below the lake surface as the cloud is above.
2. Equate the two expressions for the horizontal distance.

Answer: Let width = w and height = h. h = w tan 60° and h = (w + 40) tan 30°, so 3w = w + 40 and w = 20 m. Height = 20√3 m ≈ 34.64 m.

Teacher's Note:
1. Form two equations with the same height.
2. Solve for the river width first.