### APPLICATIONS OF TRIGONOMETRY

Important questions

4 Marks Questions

1. A person standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 20 m. From the bank, finds the angle to be 30°. Find the height of the tree and the breadth of the river.

Sol: Let width of the river AB = x m Height of the tree BC = h m

Angle of elevation at A is 60°

'D' be the position of the person after retreating 20 m from the bank So AD = 20 m and angle at D is 30°  3x − x = 20

⇒  x = 10 m

from (1), h = 10 m

∴ Hence width of the river AB = 10 m, height of the river BC = 10 m

2. The shadow of a vertical tower on level ground increases by 10 m. Which the altitude of the sun changes from angle of elevation 45° to 30°. Find the height of the tower.

Sol: Let AB be the tower, AC and AD be its shadows when the angles of elevation of the sun are 45° and 30° respectively.

Then CD = 10 m, Let AB = h and AC = x m Then CD = 10 m, Let AB = h and AC = x m

= 5(1.732 + 1) = 5(2.732) = 13.660
∴ Height of the tower AB = 13.66 m

3. An aeroplane at an altitude of 1200 m. Find 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° respectively. Find the distance between the two ships.

Sol: Let the aeroplane be at B and Let two ships be at C and D such that their angles of depression from B are 30° and 60° respectively we have AB = 1200 m,

Let

AC = x and CD = y  ⇒  y = 1200 ⇒  y = 1200 − 400 = 800 ⇒  y = 800 × 1.732 = 8 × 173.2

y = 1385.6 m

∴ Hence, the distance between two ships = 1385.6 m

4. Two pillers of equal height and on either side of a road, which is 100 m wide. The angels of elevation of the top of the pillers are 60° and 30° at a point on the road between the pillers. Find the position of the point between the pillers and height of each piller.

Sol: Let AB and CD be two pillers, each of height 'h' meters.

Let P be a point on the road such that AP = x meters

Then CP = (100 − x) m, It is given that ∠APB = 60° and ∠CPD = 30°

In ΔPAB, we have ∴ Hence, the required point is at a distance of 25 meters from the first piller, 75 m from second piller and the height of the piller is 43.3 m.

5. From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.

Sol: Height of cell tower BE = BC + CE = h m Height of the building AD = BC = x m

Distance between foot of the tower and foot of the building AB = 7 m

CE = BE − BC = (h − x) m

Angle of elevation from D = ∠CDE = 60°

Angle of depression from D= ∠CDB = 45°

∴ ∠ABD = ∠CDB = 45° (∵ Alternate angles) from (1) and (2) we hav
h − 7 = 7 ⇒ h = 7 + 7 h = 7(1 + ) ⇒ h = 7(2.732)
= 19.124
∴ Height of cell tower h = 19.124 m

6. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.

Sol: Height of tower AB = h
Distance from 1st tower to observation point AC = 4 m
Distance from 2nd tower to observation point AD = 9 m
∠ACB = x, then ∠ADB = 90 − x
In ΔABC, tan x  from (1) and (2) we hav ∴ Height of the tower h = 6 m

7. A wire of length 18 m. had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut?

Sol: Height of electric pole AD = 'h' m Length of the wire which tied by making 30°
angle with ground AB = 18 m
Length of wire which tied by making 60°
angle with ground AC = 'x' m
In Δ ABD,  Total length of the wire = 18 m
If 10.392 m wire is used then length of remaining wire
= 18 − 10.392 = 7.608 m

8. An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?

Sol: Height of the pole AB = 9 m Length of the ladder = CD
Height of electric pole to do repair work = AC
AC = 9 − 1.8 = 7.2 m
Distance between foot of the ladder to foot of the pole = AD = 4.8(1.732) = 8.3136 m = 2.4(1.732) = 4.1562 m
∴ The distance between foot of the ladder and foot of the pole = 4.1562 m.

2 MARKS QUESTIONS

1. Atree breaks due to storm and broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between foot of the tree and top of the tree on the ground is 6 m. Find the height of the tree before falling down. Sol: Height of broken part of the tree = AB
remaining part of broken tree = AC
Angle of elevation of broken part of tree with ground = ∠ACB = 30°
Distance between foot of the tree and top of the tree on ground = BC = 6 m.  Height of the tree before broken = AB + AC 2. Length of the shadow of a 15 m high pole is 5 m at 7 AM. Then, what is the angle of elevation of the sun rays with the ground at that time?

Sol: Height of the pole AB = 15 m Length of shadow of pole BC = 5 m.
Let the angle of elevation of sun rays with the ground is 'θ' ∴ θ = 60°
∴ The angle of elevation = 60

3. The boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due of stream of the river and travels a distance of 600 m to reach another side of the river. What is the width of the river?

Sol: Let width of the river = AB Angle at the river = 60°
travel of boat from A to C = AC = 600 m ∴ Width of the river = 300 m

1 MARK QUESTIONS

1. A tower stands vertically on the ground. From a point which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?

Sol: Let the height of the tower = AB Distance between foot of the tower and observation point C is BC = 15 m
Angle of elevation at C is 45° ∴ Height of the tower = 15 m

2. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.

Sol: Let the height of the tower = AB Distance between foot of the tower and observation point C is BC = 30 m
Angle of elevation at C is 30° 3. Draw suitable diagram for the following situation.

"The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m." Sol: Let Height of the first tower = AB
Height of second tower = CD = 60 m
Distance between two towers = AC = 140 m
The angle of elevation of top B of tower AB as seen from D is 30°

4. Draw suitable diagram for the following situation. "As observed from the top of a light house, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°."

Sol: Let A and B be the two positions of the ship
Let D be the observing point, top of the light house CD = 100 m.
The angles of depression from
D of A and B are 30° and 45° respectively.

Writer: T.S.V.S. Suryanarayana Murthy

Posted Date : 05-11-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.