EXERCISE 9.1
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 ).
Let AB be the vertical pole Ac be 20 m long rope tied to point C.
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.
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 be the broken part of the tree.
∴ Total height of the tree = AB+AC
In right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
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 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
A/q,
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.