- 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 °.
It can be observed from the figure that AB is the pole.
AB/AC = sin 30°
AB/20 = 1/2
AB = 20/2 = 10
Therefore, the height of the pole is 10 m.
- 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 the original tree. Due to storm, it was broken into two parts.
The broken part AB is making 30° with the ground.
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
Height of the tree = A’B + BC
= (8/√3 + 16/√3)m = 24/√3 m = 8√3m
Hence, the height of the tree is 8√3 m
- 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, where as for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60 ° to the ground. What should be the length of the slide in each case?
It can be observed that AC and PR are the slides for younger and elder children respectively.
AB/AC = sin30°
1.5/AC = 1/2
AC = 3 m
PQ/PR = sin60°
3/PR = √3/2
PR = 6/√3 = 2√3 m
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 be the tower and the angle of elevation from point C is 30°.
AB/AC = tan30°
AB/30 = 1/√3
AB = 30/√3 = 10√3m
Therefore, the height of the tower is 10√3 m.
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 be the kite and the string is tied to point P on the ground.
KL/KP = sin30°
60/KP = √3/2
KP = 120/√3 = 40√3m
Hence, the length of the string is 40√3 m
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 the boy was standing at point S initially. He walked towards the building and reached at point T.
PR = PQ − RQ = (30 − 1.5) m = 28.5 m = 57/2 m
PR/AR = tan30°
57/AR = 1/√3
AR = (57/2 √3) m
PR/BR = tan60°
57/2 BR = √3
BR = 56/2√3 = m
ST = AB = AR – BR = (57/2 √3 – 19√3/2)m = 38/2 √3 m = 19 √3 m
Hence, he walked 19√3 m towards the building.
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 BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.
BC/CD = tan45°
20/CD = 1
CD = 20 m
AC/CD = tan60°
AB+BC/CD = √3
AB = (20√3 – 20)m
= 20(√3 – 1)m
Therefore, the height of the transmission tower is 20(√3 − 1) m.
8. A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of 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 AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.
BC/CD = tan45°
BC/CD = 1
CD = BC
AB+BC/CD = tan60°
AB+BC/CD = √3
1.6 + BC = BC√3
BC(√3 – 1)= 1.6
BC = (1.6)( √3+1)/(√3-1)( √3+1)
= (1.6)( √3+1)/(√3)2– (1)2
= 1.6(√3+1)/2 = 0.8(√3+1)
Therefore, the height of the pedestal is 0.8(√3 + 1) m.
- 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 the building and CD be the tower. In ∆CDB,
CD/BD = tan60°
50/BD = √3
BD = 50/√3 m
AB/BD = tan30°
AB/CD = 50/√3 X 1/√3 = 50/3 = 16 2/3
Therefore, the height of the building is 16 2/3 m.