Extra Practice Questions
Applications of Trigonometry (R.D. Sharma Level)
1. An aeroplane at an altitude of 1200 meters finds that two ships are sailing towards it in the same direction. The angles of depression of the ships as observed from the aeroplane are 60° and 30°. Find the distance between the two ships.
Let the aeroplane be at A, 1200 m high. Let the ships be at C and D.
For the nearer ship (C) with depression angle 60°: tan 60° = 1200/BC => √3 = 1200/BC => BC = 1200/√3 = 400√3 m.
For the farther ship (D) with depression angle 30°: tan 30° = 1200/BD => 1/√3 = 1200/BD => BD = 1200√3 m.
Distance between ships = CD = BD – BC = 1200√3 – 400√3 = 800√3 m.
Answer: The distance between the ships is 800√3 m.
2. A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower?
Let tower height be h. Let car be at C (angle 30°) then at D (angle 45°).
In ΔABD: tan 45° = h/BD => 1 = h/BD => h = BD.
In ΔABC: tan 30° = h/BC = h/(BD+CD) => 1/√3 = h/(h+CD) => h+CD = h√3 => CD = h(√3 – 1).
Time to travel distance CD = h(√3 – 1) is 12 minutes.
Speed = Distance/Time = h(√3 – 1)/12.
Time to travel distance BD = h.
Time = Distance/Speed = h / [h(√3 – 1)/12] = 12/(√3 – 1) = 6(√3 + 1) minutes.
Using √3 ≈ 1.732, Time ≈ 6(2.732) = 16.392 minutes.
Answer: Approx. 16.39 minutes (or 16 minutes and 24 seconds).
3. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Let tower height be h. Let shadow length be x when angle is 60°, and x+40 when angle is 30°.
Case 1 (angle 60°): tan 60° = h/x => √3 = h/x => x = h/√3.
Case 2 (angle 30°): tan 30° = h/(x+40) => 1/√3 = h/(x+40) => x+40 = h√3.
Substitute x: (h/√3) + 40 = h√3 => 40 = h√3 – h/√3 = h(3-1)/√3 = 2h/√3.
80 = 2h/√3 => h = 40√3 / √3 = 20√3 m.
Answer: The height of the tower is 20√3 m.
… (17 more questions follow)
4. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is h tanα / (tanβ – tanα).
Let tower height be H, flagstaff height be h. Let distance to the point be x.
Elevation to bottom (top of tower): tan α = H/x => x = H/tanα.
Elevation to top (top of flagstaff): tan β = (H+h)/x => x = (H+h)/tanβ.
Equate x: H/tanα = (H+h)/tanβ => H tanβ = H tanα + h tanα.
H tanβ – H tanα = h tanα => H(tanβ – tanα) = h tanα.
H = h tanα / (tanβ – tanα). (Proved)
5. From the top of a cliff 150 m high, the angles of depression of two boats on the same side are found to be 45° and 60°. Find the distance between the boats.
Let cliff height be AB=150m. Let boats be at C (60°) and D (45°).
For boat at C: tan 60° = 150/BC => √3 = 150/BC => BC = 150/√3 = 50√3 m.
For boat at D: tan 45° = 150/BD => 1 = 150/BD => BD = 150 m.
Distance CD = BD – BC = 150 – 50√3 = 50(3 – √3) m.
Answer: The distance is 50(3 – √3) m.
6. The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.
Let A be the point of observation, 60m above lake level B. Let C be the cloud and C’ be its reflection. Height of cloud above lake = h. So, CB = C’B = h.
In the upper triangle: tan 30° = (h-60)/AB => 1/√3 = (h-60)/AB => AB = (h-60)√3.
In the lower triangle (with reflection): tan 60° = (h+60)/AB => √3 = (h+60)/AB => AB = (h+60)/√3.
Equate AB: (h-60)√3 = (h+60)/√3 => 3(h-60) = h+60 => 3h-180=h+60 => 2h=240 => h=120 m.
Answer: The height of the cloud from the lake is 120 m.
7. A round balloon of radius r subtends an angle α at the eye of the observer while the angle of elevation of its centre is β. Prove that the height of the centre of the balloon is r sinβ cosec(α/2).
Let O be the observer’s eye, C be the balloon’s centre, and P be a tangent point on the balloon from O. Let H be the height of the centre C. Then OC is the hypotenuse of a right triangle with angle β.
In the right triangle formed by O, C, and the horizontal line, sinβ = H/OC => OC = H/sinβ.
In the right triangle OPC (where CP=r is the radius), the angle ∠POC = α/2.
sin(α/2) = CP/OC = r/OC => OC = r/sin(α/2) = r cosec(α/2).
Equate the two expressions for OC: H/sinβ = r cosec(α/2).
H = r sinβ cosec(α/2). (Proved)
8. A man on a bridge 25m high, observes the angles of depression of two boats on the opposite sides of the bridge to be 45° and 30°. Find the distance between the boats.
Let bridge height be h=25m. Let boats be at C (45°) and D (30°).
Distance to boat C: tan 45° = 25/BC => 1 = 25/BC => BC = 25 m.
Distance to boat D: tan 30° = 25/BD => 1/√3 = 25/BD => BD = 25√3 m.
Total distance = BC + BD = 25 + 25√3 = 25(1 + √3) m.
Answer: The distance between the boats is 25(1 + √3) m.
9. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.
Height h=1 km. Let initial horizontal distance be x, final be y.
tan 60° = 1/x => √3 = 1/x => x = 1/√3 km.
tan 30° = 1/y => 1/√3 = 1/y => y = √3 km.
Distance travelled = y-x = √3 – 1/√3 = 2/√3 km.
Time = 10 sec = 10/3600 hr = 1/360 hr.
Speed = Distance/Time = (2/√3) / (1/360) = 720/√3 = 240√3 km/hr.
Answer: The speed is 240√3 km/hr.
10. From a window 15 metres high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 15(1 + 1/√3) metres.
Let window be at A, 15m high. Let the opposite house be CD. Let street width be x.
Depression to foot (D): tan 45° = 15/x => 1 = 15/x => x = 15 m.
Elevation to top (C): tan 30° = CE/x where CE is height above window level.
1/√3 = CE/15 => CE = 15/√3 m.
Total height of house = ED + CE = 15 + 15/√3 = 15(1 + 1/√3) m.
(Proved)
11. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Let man be at A, 10m high. Let hill be CD. Let horizontal distance be x.
Depression to base (D): tan 30° = 10/x => 1/√3 = 10/x => x = 10√3 m.
Elevation to top (C): tan 60° = CE/x => √3 = CE/(10√3) => CE = 30 m.
Height of hill = ED + CE = 10 + 30 = 40 m.
Answer: Distance is 10√3 m, Height is 40 m.
12. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 m, find the speed of the jet plane.
Height h=1500√3 m. Let initial horizontal distance be x, final be y.
tan 60° = h/x => √3 = 1500√3/x => x=1500 m.
tan 30° = h/y => 1/√3 = 1500√3/y => y=4500 m.
Distance covered = y-x = 4500 – 1500 = 3000 m. Time = 15 s.
Speed = Distance/Time = 3000/15 = 200 m/s.
In km/hr: 200 × (18/5) = 720 km/hr.
Answer: The speed is 200 m/s or 720 km/hr.
13. At the foot of a mountain the elevation of its summit is 45°; after ascending 1000m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.
Let height be H. Let initial point be C. Let new point be D after ascending 1000m up a 30° slope.
Vertical rise from C to D = 1000 sin 30° = 500m.
Horizontal distance covered = 1000 cos 30° = 500√3m.
From C: tan 45° = H/BC => BC = H.
From D: tan 60° = (H-500)/(BC-500√3) => √3 = (H-500)/(H-500√3).
H√3 – 1500 = H-500 => H(√3 – 1) = 1000 => H = 1000/(√3 – 1) = 500(√3 + 1) m.
Answer: The height is 500(√3 + 1) m.
14. If the angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the height of the cloud is h(tanβ + tanα) / (tanβ – tanα).
Let cloud height from lake be H. Observer is at height h. Horizontal distance is x.
tan α = (H-h)/x => x = (H-h)/tanα.
tan β = (H+h)/x => x = (H+h)/tanβ.
Equate x: (H-h)/tanα = (H+h)/tanβ.
H tanβ – h tanβ = H tanα + h tanα.
H(tanβ – tanα) = h(tanβ + tanα).
H = h(tanβ + tanα) / (tanβ – tanα). (Proved)
15. A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.
First boy: Height of kite h₁ = 100 sin 30° = 50 m. Horizontal distance x₁ = 100 cos 30° = 50√3 m.
Second boy is on a 10m building. For kites to meet, they must be at the same height. Height of second kite from ground must be 50m.
Height of second kite from boy = 50 – 10 = 40 m.
Let L be the length of the second string. Angle is 45°.
sin 45° = 40/L => 1/√2 = 40/L => L = 40√2 m.
Answer: The length of the string is 40√2 m.
16. The angle of elevation of the top of a tower from a point A due south of the tower is α and from a point B due east of the tower is β. If AB = d, show that the height of the tower is d / √(cot²α + cot²β).
Let tower be at O, height h. Point A is south, B is east. ΔOAB is a right triangle with AB=d.
From A: tan α = h/OA => OA = h/tanα = h cotα.
From B: tan β = h/OB => OB = h/tanβ = h cotβ.
In ΔOAB, by Pythagoras theorem: OA² + OB² = AB².
(h cotα)² + (h cotβ)² = d² => h²(cot²α + cot²β) = d².
h² = d² / (cot²α + cot²β) => h = d / √(cot²α + cot²β). (Proved)
17. A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through a distance ‘a’, so that it slides a distance ‘b’ down the wall making an angle β with the horizontal. Show that a/b = (cosα – cosβ) / (sinβ – sinα).
Let ladder length be L. Initially, let it touch the wall at height y and be at distance x from the wall.
Initial state: y = L sinα, x = L cosα.
Final state: New height y’ = y-b. New distance x’ = x+a.
y-b = L sinβ => b = y – L sinβ = L sinα – L sinβ = L(sinα – sinβ).
x+a = L cosβ => a = L cosβ – x = L cosβ – L cosα = L(cosβ – cosα).
The question has a slight typo. It should be a/b = (cosβ – cosα)/(sinα – sinβ). Let’s use that.
Ratio a/b = [L(cosβ – cosα)] / [L(sinα – sinβ)] = (cosβ – cosα)/(sinα – sinβ). (Proved as per correction)
18. The angles of elevation of the top of a tower from the top and bottom of a 100 m high cliff are 30° and 60° respectively. Find the height of the tower.
Let cliff be AB=100m, tower be CD. Let horizontal distance be x. Cliff is higher than tower in this case.
From bottom of cliff (B): tan 60° = CD/x => √3 = CD/x => x = CD/√3.
From top of cliff (A): tan 30° = (CD-100)/x => 1/√3 = (CD-100)/x => x = √3(CD-100).
Equate x: CD/√3 = √3(CD-100) => CD = 3(CD-100) => CD = 3CD – 300 => 2CD = 300 => CD=150m.
Wait, the diagram should have tower higher than cliff. Let’s re-read. Let’s assume Tower is higher.
Let tower be AB=h, cliff CD=100m. From D: tan 60° = h/BD. From C: tan 30° = (h-100)/BD.
h/tan 60° = (h-100)/tan 30° => h/√3 = (h-100)/(1/√3) => h/√3 = √3(h-100).
h = 3(h-100) => h = 3h-300 => 2h = 300 => h=150m.
Answer: The height of the tower is 150 m.
19. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°. If the bridge is at a height of 3 m from the banks, find the width of the river.
Let bridge height be h=3m. Let the point on the bridge be P. Let the banks be A and B.
Distance to bank A: tan 45° = 3/x₁ => x₁ = 3 m.
Distance to bank B: tan 30° = 3/x₂ => 1/√3 = 3/x₂ => x₂ = 3√3 m.
Width of river = x₁ + x₂ = 3 + 3√3 = 3(1 + √3) m.
Answer: The width of the river is 3(1 + √3) m.
20. Two men on either side of a 75 m high building and in line with the base of the building observe the angles of elevation of the top of the building as 30° and 60°. Find the distance between the two men.
Let building height be h=75m.
Let distance of first man be x₁. tan 30° = 75/x₁ => 1/√3 = 75/x₁ => x₁ = 75√3 m.
Let distance of second man be x₂. tan 60° = 75/x₂ => √3 = 75/x₂ => x₂ = 75/√3 = 25√3 m.
Total distance = x₁ + x₂ = 75√3 + 25√3 = 100√3 m.
Answer: The distance between the men is 100√3 m.
