TOPICS

Exercise - 9.1

Some Applications of Trigonometry

**Question-1 :-** A circus artist is climbing a 20 m long rope, which is
tightly stretched and tied from the top of a vertical
pole to the ground. Find the height of the pole, if
the angle made by the rope with the ground level is
30° (see Figure).

Here, AB is the Pole. Length of rope (AC) = 20 m Height of the pole (AB) = ? In Δ ABC, AB/AC = sin 30° AB/20 = 1/2 AB = 20/2 AB = 10 m Therefor, Height of the pole is 10 m.

**Question-2 :-** A tree breaks due to storm and the broken part
bends so that the top of the tree touches the ground
making an angle 30° with it. The distance between
the foot of the tree to the point where the top
touches the ground is 8 m. Find the height of the
tree.

Let AC was a original tree. Length of Break part of tree (A'C) = 8 m Total Height of the tree (AC) = ? In Δ A'BC, BC/A'C = tan 30° BC/8 = 1/√3 BC = 8/√3 m A'C/A'B = cos 30° 8/A'B = √3/2 A'B = 16/√3 m Total Heght of tree (AC) = A'B + BC = 16/√3 + 8/√3 = 24/√3 = 8√3 m Therefor, Height of the tree is 8√3 m.

**Question-3 :-** A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have
a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case?

Let AC and PR be the slides of Younger and Elder Children respectively. AB = 1.5 m, PQ = 3 m In Δ ABC, AB/AC = sin 30° 1.5/AC = 1/2 AC = 3.0 m Therefore, length of younger slides is 3 m. In Δ PQR, PQ/PR = sin 60° 3/PR = √3/2 PR = 6/√3 = 2√3 m Therefore, length of elder slides is 2√3 m.

**Question-4 :-** 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.

Let AB the tower and Distance BC = 30 m Angle of elevation = 30° Height of the tower (AB) = ? In Δ ABC, AB/BC = tan 30° AB/30 = 1/√3 AB = 30/√3 = 10√3 m Therefore, the height of tower is 10√3 m.

**Question-5 :-** A kite is flying at a height of 60 m above the ground. The string attached to the kite is
temporarily tied to a point on the ground. The inclination of the string with the ground
is 60°. Find the length of the string, assuming that there is no slack in the string.

Let K is a Kite. Height of kite from the ground (LK) = 60m Length of string (KP) = ? In Δ KLP, KL/KP = sin 60° 60/KP = √3/2 KP = 120/√3 = 40√3 m Therefore, the Length of string is 40√3 m.

**Question-6 :-** A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance he walked towards the building.

Let a boy was standing at point S initially. He walked towards the building and reached at T point. Height of the Boy = 1.5 m Height of the building (PQ) = 30 m So, PR = PQ - RQ = 30 - 1.5 = 28.5 = 57/2 m Distance from walked towards the building (ST) = ? In Δ PAR, PR/AR = tan 30° 57/2AR = 1/√3 AR = (57√3)/2 m In Δ PRB, PR/BR = tan 60° 57/2BR = √3 BR = 57/(2√3) = (19√3)/2 m Now, ST = AB = AR - BR = (57√3)/2 - (19√3)/2 = (38√3)/2 = 19√3 m Therefore, the Distance from walked towards the building is 19√3 m.

**Question-7 :-** From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.
Find the height of the tower.

Let AB be the Transmission Tower. Height of building (BC) = 20 m AC = AB + BC Height of Transmission Tower (AB) = ? In Δ BCD, BC/CD = tan 45° 20/CD = 1 CD = 20 m In Δ ACD, AC/CD = tan 60° (AB + BC)/BC = √3 AB + BC = √3 BC √3 BC - BC = AB BC (√3 - 1) = AB 20(√3 - 1) = AB AB = 20(√3 - 1) m Therefore, the height of Transmission tower is 20(√3 - 1) m.

**Question-8 :-** A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the
angle of elevation of the top of the statue is 60° and from the same point the angle of
elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Let the height of pedestal is BC. Height of statue (AB) = 1.6 m AC = AB + BC Height of pedestal (BC) = ? In Δ BCD, BC/CD = tan 45° BC/CD = 1 BC = CD In Δ ACD, AC/CD = tan 60° (AB + BC)/BC = √3 1.6 + BC = √3 BC √3 BC - BC = 1.6 BC (√3 - 1) = 1.6 BC = 1.6/(√3 - 1) By rationalising of Denominator BC = 1.6(√3 + 1)/[(√3 - 1)(√3 + 1)] BC = 1.6(√3 + 1)/2 BC = 0.8(√3 + 1) m Therefore, the height of the pedestal is 0.8(√3 + 1) m.

**Question-9 :-** The angle of elevation of the top of a building from the foot of the tower is 30° and the
angle of elevation of the top of the tower from the foot of the building is 60°. If the tower
is 50 m high, find the height of the building.

Let AB be a Building. Height of tower (CD) = 50 m Height of the Building (AB) = ? In Δ CDB, CD/BD = tan 60° 50/BD = √3 BD = 50/√3 m In Δ ABD, AB/BD = tan 30° AB/BD = 1/√3 AB = BD/√3 AB = 50/(√3 x √3) AB = 50/3 m Therefore, the height of the building is 50/3 m.

**Question-10 :-** Two poles of equal heights are standing opposite each other on either side of the road,
which is 80 m wide. From a point between them on the road, the angles of elevation of
the top of the poles are 60° and 30°, respectively. Find the height of the poles and the
distances of the point from the poles.

Let AB and CD are be the two equal poles and BO and Do distances of the point from the poles. Distance between both poles (BD) = 80 m Height of poles = ? Distances of the point from the poles = ? In Δ ABO, AB/BO = tan 60° AB/BO = √3 BO = AB/√3 In Δ CDO, CD/DO = tan 30° CD/DO = 1/√3 CD/(80 - BO) = 1/√3 CD √3 = 80 - BO CD √3 = 80 - AB/√3 CD √3 + AB/√3 = 80 By equal Poles, CD = AB CD √3 + CD/√3 = 80 CD(√3 + 1/√3) = 80 CD(4/√3) = 80 CD = (80√3)/4 CD = 20√3 m or AB = 20√3 m BO = AB/√3 = (20√3)√3 = 20 m DO = BD - BO = 80 - 20 = 60 m Therefore, the height of Poles are 20√3 m and the point is 20 m and 60 m far from these poles.

**Question-11 :-** A TV tower stands vertically on a bank
of a canal. From a point on the other
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m
away from this point on the line joing
this point to the foot of the tower, the
angle of elevation of the top of the
tower is 30° (see Figure). Find the
height of the tower and the width of
the canal.

Let AB be a TV Tower and BC be a Canal. The height of the tower (AB) and the width of the canal (BC) = ? In Δ ABC, AB/BC = tan 60° AB/BC = √3 BC = AB/√3 In Δ ABD, AB/BD = tan 30° AB/(BC + CD) = 1/√3 AB/(AB/√3 + 20) = 1/√3 (√3 AB)/(AB + 20√3) = 1/√3 3AB = AB + 20√3 3AB - AB = 20√3 2AB = 20√3 AB = (20√3)/2 AB = 10√3 m BC = AB/√3 = (10√3)/√3 = 10 m Therefore, the height of tower is 10√3 m and width of canal is 10 m.

**Question-12 :-** From the top of a 7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Let AB be a building and CD be a cable tower. Height of building (AB) = 7 m Height of tower = ? In Δ ABD, AB/BD = tan 45° 7/BD = 1 BD = 7 m In Δ ACE, AE = BD = 7m CE/AE = tan 60° CE/7 = √3 CE = 7√3 m CD = CE + ED = 7√3 + 7 = 7(√3 + 1) m Therefore, the height of cable tower is 7(√3 + 1) m. Therefore, the height of tower is 10√3 m.

**Question-13 :-** As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships.

Let AB be a light house and C and D are be the two ships. Height of light house (AB) = 75 m The distance between the two ships (CD) = ? In Δ ABC, AB/BC = tan 45° 75/BC = 1 BC = 75 m In Δ ABD, AB/BD = tan 30° 75/(BC + CD) = 1/√3 BC + CD = 75√3 75 + CD = 75√3 CD = 75√3 - 75 CD = 75(√3 - 1) m Therefore, The distance between the two ships 75(√3 - 1) m.

**Question-14 :-** A 1.2 m tall girl spots a balloon moving
with the wind in a horizontal line at a
height of 88.2 m from the ground. The
angle of elevation of the balloon from
the eyes of the girl at any instant is
60°. After some time, the angle of
elevation reduces to 30° (see Figure).
Find the distance travelled by the
balloon during the interval.

Let A and B are be the initial point and final point of baloon from the ground. Height of baloon from the ground (BH) or (AF) = 88.2 m Height of girl (CD) = 1.2m Distance travelled by the baloon (EG) = ? In Δ ACE, AE/CE = tan 60° (AF - EF)/CE = √3 (88.2 - 1.2)/CE = √3 87/CE = √3 CE = 87/√3 = 29√3 m In Δ BCG, BG/CG = tan 30° (BH - GH)/CG = 1/√3 (88.2 - 1.2)/CG = 1/√3 87/CG = 1/√3 CG = 87√3 m Therefore, Distance travelled by baloon (EG) = CG - CE = 87√3 - 29√3 = 58√3 m

**Question-15 :-** A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Let AB be a tower and C and D are be the initial and final positions of car. Time taken = ? In Δ ABD, AB/BD = tan 60° AB/BD = √3 BD = AB/√3 In Δ ABC, AB/BC = tan 30° AB/(BD + CD) = 1/√3 AB √3 = BD + CD AB √3 = AB/√3 + CD CD = AB √3 - AB/√3 CD = 2AB/√3 Therefore, Time taken by the car to travel distance CD (2AB/√3) = 6 seconds Time taken by the car to travel distance DB (AB/√3) = 6/(2AB/√3) x (AB/√3) = 6/2 = 3 seconds

**Question-16 :-** The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary.
Prove that the height of the tower is 6 m.

Let AQ be a tower and Base RQ and SQ are be the 4m and 9m respectively. The angles are complementary. If one angle is θ and another is (90° - θ) In Δ AQR, AQ/QR = tan θ AQ/4 = tan θ .....(1) In Δ AQS, AQ/QS = tan(90 - θ) AQ/9 = cot θ .....(2) Multipying eq (1) and (2) AQ/4 x AQ/9 = tan θ . cot θ AQ/36 = tan θ . 1/tan θ AQ²/36 = 1 AQ² = 36 AQ = √36 AQ = 6 Therefore, the height of tower is 6 m (Hence Proved).

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