10th SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 7.2 PART 6

This page 10th samacheer kalvi math solution for exercise 7.2 part 6 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 6

Question 15 :

The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be 30° and 60°. The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is 4000 km, find the distance between the satellite and earth. (√3 = 1.732)

Solution :

BC = x m

In triangle ABC :

∠ACB = 60°

tan θ = opposite side/Adjacent side

tan 60° = AB/BC

√3 = AB/BC

AB = √3BC

AB = x√3 ----(1)

In triangle ABD :

∠ADB = 30°

tan θ = opposite side/Adjacent side

tan 30° = AB/BD

1/√3 = AB/(BC + CD)

1/√3 = AB/(x + 4000)

AB = (x + 4000)/√3 ---->(2)

(1) = (2)

x√3 = (x + 4000)/√3

3 x = x + 4000

3x - x = 4000

2x = 4000

x = 2000

distance between the satellite and earth(AB) = x√3

= 2000 (1.732) = 3464 km

Question 16 :

From the top of a tower of height 60 m, the angles of depression of the top and the bottom of a building are observed to be 30° and 60° respectively. Find the height of the building.

Solution :

From the given information, we can draw a rough diagram

In triangle ABC :

Let AB = x and BD = 60 - x

∠ACB = 30°

tan θ = opposite side/Adjacent side

tan 30° = AB/BC

1/√3 = x/BC

BC = x √3 ----(1)

In triangle ADE :

∠AED = 60°

tan θ = opposite side/Adjacent side

tan 60° = AD/DE

√3 = 60/DE

DE = 60/√3----(2)

(1) = (2)

x √3 = 60/√3

3x = 60

x = 20 m

CE = 60 - 20 = 40 m

Height of the building = 40 m.

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