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1. A man is walking along a straight road. He notices the top of a tower subtending an angle A = 60^{o} with the ground at the point where he is standing. If the height of the tower is h = 20 m, then what is the distance (in meters) of the man from the tower? Answer: 11.55
Solution: 
Let BC represent the tower with height h = 20 m, and A represent the point where the man is standing. AB = d denotes the distance of the man from tower. The angle subtended by the tower is A = 60^{o}.
From trigonometry, tan A = tan 60^{o} = h / d =  √3 
So d = 20 / √3 s m.
Hence the distance of the man from the tower is 11.55 m. 
2. A little boy is flying a kite. The string of the kite makes an angle of 30^{o} with the ground. If the height of the kite is h = 15 m, find the length (in meters) of the string that the boy has used. Answer: 30
Solution: 
If the kite is at C and the boy is at A, then AC = l represents the length of the string and BC = h represents the height of the kite.
From the figure, sin A = sin 30^{o} = h / l = 1 / 2. Hence the length of the string used by the little boy is l = 2 h = 2 (15) = 30 m. 
3. Two towers face each other separated by a distance d = 20 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 60^{o} and that of the top is 30^{o}. What is the height (in meters) of the second tower?
Answer: 23.09
Solution: 
The first tower AB and the second tower CD are depicted in the figure on the left.
First consider the triangle BAC. Angle C = 60^{o}.
tan BCA = tan 60^{o} = AB / AC.
This gives AB = d tan 60^{o}.
Similarly for the triangle BED, BE = d tan 30^{o}.
Now height of the second tower CD = AB − BE
= d (tan 60^{o} − tan 30^{o})
= 20 (√3 − 1/ √3) = 20 × 2 / √3 = 23.09 m. 
4. A ship of height h = 15 m is sighted from a lighthouse. From the top of the lighthouse, the angle of depression to the top of the mast and the base of the ship equal 30^{o} and 45^{o} respectively. How far is the ship from the lighthouse (in meters)?
Answer: 35.49
Solution:
Let AB represent the lighthouse and CD represent the ship. From the figure, tan BCA = tan 45^{o} = AB / AC.
Similarly for the triangle BED, tan BDE = tan 30^{o} = BE / ED.
Now, AC = ED = d.
Height of the ship = CD
= AB − BE = d (tan 45^{o} − tan 30^{o}) = 15 m.
Thus distance of the ship from the lighthouse d = 15 / (1 − 1 / √3 ) = 35.49 m  
5. Two men on opposite sides of a TV tower of height 28 m notice the angle of elevation of the top of this tower to be 45^{o} and 60^{o} respectively. Find the distance (in meters) between the two men. Answer: 44.17
Solution:
The situation is depicted in the figure with CD representing the tower and AB being the distance between the two men.
For triangle ACD,
tan A = tan 60^{o} = CD / AD.
Similarly for triangle BCD,
tan B = tan 45^{o} = CD / DB.
The distance between the two men is AB = AD + DB
= (CD / tan 60^{o}) + (CD / tan 45^{o})
= (28 / √3) + (28 / 1) = 44.17 m.  
6. Two men on the same side of a tall building notice the angle of elevation to the top of the building to be 30^{o} and 60^{o} respectively. If the height of the building is known to be h =50 m, find the distance (in meters) between the two men. Answer: 57.73
Solution: In the figure, A and B represent the two men and CD the tall building.
tan A = tan 30^{o} = DC / AC = h / AC; and
tan B = tan 60^{o} = DC / BC = h / BC.
Now the distance between the men is AB
= x = AC − BC = (h / tan 30^{o}) − (h / tan 60^{o})
= (50 √3 ) − (50 / √3 ) = 57.73 m.  
7. A pole of height h = 60 ft has a shadow of length l = 34.64 ft at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time. Answer: 60
Solution: 
In the figure, BC represents the pole and AB its shadow.
tan A = BC / AB
= h / l = 60 / 34.64 = 1.732
From trigonometric tables, we note that
tan A = 1.732 for A =60^{o}.
Hence the angle of elevation of the sun at this point of time is 60^{o}.

8. You are stationed at a radar base and you observe an unidentified plane at an altitude h = 4000 m flying towards your radar base at an angle of elevation = 30^{o}. After exactly one minute, your radar sweep reveals that the plane is now at an angle of elevation = 60^{o} maintaining the same altitude. What is the speed (in m/s) of the plane? Answer: 76.98
Solution:  In the figure, the radar base is at point A. The plane is at point D in the first sweep and at point E in the second sweep. The distance it covers in the one minute interval is DE.
From the figure,
tan DAC = tan 30^{o} = DC / AC = h / AC.
Similarly,
tan EAB = tan 60^{o} = EB / AB = h / AB.
Distance covered by the plane in one minute = DE = AC − AB
= (h / tan 30^{o}) − (h / tan 60^{o})
= (4000 √3) − (4000 / √3 ) = 4618.80 m.
The velocity of the plane is given by V = distance covered / time taken
= DE / 60 = 76.98 m/s. 
Try the Quiz : Practice Exercise for Trigonometry Module 3 : Word Problems on Heights and Distances
