# 10th SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 7.2 part 5

This page 10th samacheer kalvi math solution for exercise 7.2 part 5 is going to provide you solution for every problems that you find in the exercise no 7.2

## 10th Samacheer Kalvi Math Solution for Exercise 7.2 part 5

Question 13 :

A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height . The angle of elevation of the balloon from the boy at an instant is 60°. After 2 minutes, from the same point of observation,the angle of elevation reduces to 30°. If the speed of wind is 29√3 m/min. then, find the height of the balloon from the ground level.

Solution : Distance covered by the balloon  =  BC

BC = Time x Speed ==> 2 x 29 √3 ==> 58√3 m

AB = x  then AC = x + 58√3

In triangle DAC :

∠DAC  =  30°

tan θ = opposite side/Adjacent side

tan 30°  =  DC/AC

1/√3  =  DC/(x + 58√3)

DC = (x + 58√3)/√3 ----(1)

In triangle EAB :

∠EAB  =  60°

tan θ = opposite side/Adjacent side

tan 60°  =  EB/AB

√3  =  EB/x

x√3  =  EB

EB  =  √3x ---->(2)

Since EB = DC

(1) = (2)

(x + 58√3)/√3  =  √3x

x + 58√3  =  3x

3x - x  =  58√3

2x  =  58√3

x  =   58√3/2 ==> 29√3 m

Height of the balloon from ground level EB  =  √3 x

=  29 √3 (√3)

=  29(3) ==> 87 m

Hence height of the balloon from ground level = 87 m.

Question 14 :

A straight highway leads to the foot of a tower . A man standing on the top of the tower spots a van at an angle of depression of 30°. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60°. How many more minutes will it take for the van to reach the tower?

Solution :

From the given information, we can draw a rough diagram Distance covered by the van to reach D from C = 6 minutes

time taken = x

Distance between D and C = 6x

In triangle ACB

∠ACB  =  30°

tan θ = opposite side/Adjacent side

tan 30° = AB/BC

1/√3  =  AB/(BD+DC)

1/√3  =  AB/(BD+6x)

(BD+6x)/√3 = AB ----(1)

In triangle ABD

∠ABD  =  60°

tan θ = opposite side/Adjacent side

tan 60° = AB/BD

3  =  AB/BD ==> AB  = BD3  -----(2)

(1) = (2)

(BD+6x)/√3  =  BD3

BD + 6x = BD(3)

3BD - BD = 6x

2BD = 6x

BD = 6x /2 = 3x

Here 3 represents number of minutes covered by the van and x stands for time taken.

Hence 3 more minutes will it take for the van to reach the tower.

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