10th SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 7.2 part 4

This page 10th samacheer kalvi math solution for exercise 7.2 part 4 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 4

Question 10 :

A student sitting in a classroom sees a picture on the black board at a height of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30°. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45°. Find the distance moved by the student.

Solution :

Distance moved by the student = BC

In triangle ABD :

∠ABD = 30°

tan θ = opposite side/Adjacent side

tan 30° = AD/BD

1/√3 = 1.5/BD

BD = 1.5 x √3 ==> 1.5 √3

In triangle ACD :

∠ACD = 45°

tan θ = opposite side/Adjacent side

tan 45° = AD/CD

1 = 1.5/CD

CD = 1.5

BC = BD - CD

BC = 1.5 √3 - 1.5

= 1.5 (√3 - 1) = 1.5(1.732 - 1)

= 1.5 (0.732) ==> 1.098 m

Hence the distance moved by the student is 1.098 m.

Question 11 :

A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. 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 :

From the given information, we can draw a rough diagram

AD = 28.5 m

In triangle ABD

∠ABD = 30°

tan θ = opposite side/Adjacent side

tan 30° = AD/BD

1/√3 = 28.5/BD ==> BD = 28.5 √3 --(1)

In triangle ACD

∠ABD = 60°

tan θ = opposite side/Adjacent side

tan 60° = AD/CD

√3 = 28.5/CD ==> CD = 28.5/√3

Multiplying by √3 on both numerator and denominator, we get

CD = 28.5/√3 ==> 28.5√3/3 ==> 9.5√3 -->(2)

the distance he walked towards the building = BD - CD

= 28.5 √3 - 9.5√3 ==> 19√3 m

Question 12 :

From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a Yacht and a Barge along the same line of sight .The angles of depression for the Yacht and the Barge are 45° and 30° respectively. For safety purposes the two sea vessels should be at least 300 feet apart. If they are less than 300 feet , the keeper has to sound the alarm. Does the keeper have to sound the alarm ?

Solution :

From the above picture, we have to find the value of CD.

In triangle ABC

∠ACB = 45°

sin θ = opposite side/Hypotenuse side

sin 45° = AB/BC

1/√2 = 200/BC

BC = 200 √2

In triangle ABD

∠ADB = 30°

sin 30° = AB /BD

1/2 = 200/BD

BD = 200 x 2 ==> 400

CD = BD - BC ==> 400 - 200 √2 ==> 200(2 - √2)

= 200 (2 - 1.414)

= 200(0.586) ==> 117.2 m

From this we come to know that the distance between Yacht and a Barge is less than 300 m. Hence the keeper has to sound the alarm.

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