10th SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 7.2 part 1

This page 10th samacheer kalvi math solution for exercise 7.2 part 1is 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 1

1. A ramp for unloading a moving truck, has an angle of elevation of 30°. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.

Solution:

In any right triangle, the side which is opposite to 90 degree is known as hypotenuse side, the side which is opposite to θ is known as opposite side and the remaining side is known as adjacent side.

In the given problem,we have to find the length of hypotenuse side and we know the length of opposite side.

AC = Hypotenuse side

AB = Opposite side

BC = Adjacent side

                   sin θ = opposite side/hypotenuse side

                   sin 30° = AB/AC

                      1/2 = 0.9/AC 

                      AC = 0.9 x 2 

                      AC = 1.8 m

Therefore length of ramp is 1.8 m


2. A girl of height 150 cm stands in front of a lamp-post and casts a shadow of length 150 3 cm on the ground. Find the angle of elevation of the top of the lamp-post.

Solution:

In the given problem,we have to find the angle inclined C.

AC = Hypotenuse side

AB = Opposite side = 150 cm

BC = Adjacent side = 150 3 cm

                   tan θ = opposite side/Adjacent side

                   tan θ = AB/BC

                   tan θ = 150/150 3

                   tan θ = 1/3

                        θ = 30° 


3. Suppose two insects A and B can hear each other up to a range of 2 m. The insect A is on the ground 1 m away from a wall and sees her friend B on the wall, about to be eaten by a spider. If A sounds a warning to B and if the angle of elevation of B from A is 30°, will the spider have a meal or not ? ( Assume that B escapes if she hears A calling )

Solution:

In the given problem,we have to the length of AB.

AC = Hypotenuse side

BC = Opposite side = 1 m

AC = Adjacent side

                   Sin θ = Opposite side/Hypotenuse side

                   Sin θ = BC/AB

                  sin 3 = 1/AC

                     1/2 = 1/AC

                       AC = 2 m

So spider B escapes.




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