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).Solution :-
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.Solution :-
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?Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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.Solution :-
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).