**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 °.**

Solution:

It can be observed from the figure that AB is the pole.

In ∆ABC,

^{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**.

Solution:

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.

In ∆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

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?**

Solution:

It can be observed that AC and PR are the slides for younger and elder children respectively.

In ∆ABC,

^{AB}/_{AC} = sin30°

^{1.5}/_{AC } = ^{1}/_{2}

AC = 3 m

In ∆PQR,

^{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.**

Solution:

Let AB be the tower and the angle of elevation from point C is 30°.

In ∆ABC,

^{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. **

Solution:

Let K be the kite and the string is tied to point P on the ground.

In ∆KLP,

^{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**

Solution:

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

In ∆PAR,

^{PR}/_{AR} = tan30°

^{57}/_{AR } = ^{1}/_{√3}

AR = (^{57}/_{2 }√3) m

In ∆PRB,

^{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. **

Solution:

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.

In ∆BCD,

^{BC}/_{CD} = tan45°

^{20}/_{CD } = 1

CD = 20 m

In ∆ACD,

^{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. **

Solution:

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.

In ∆BCD,

^{BC}/_{CD} = tan45°

^{BC}/_{CD} = 1

CD = BC

In ∆ACD,

^{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.**

Solution:

Let AB be the building and CD be the tower. In ∆CDB,

^{CD}/_{BD} = tan60°

^{50}/_{BD } = √3

BD = ^{50}/_{√}_{3} m

In ∆ABD,

^{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.

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