In this section, you will learn how to solve word problems in trigonometry step by step.

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slide 1:

Trigonometry Word Problems with
Solutions
In this section you will learn how to solve word problems in trigonometry step by step.
Trigonometry Word Problems with
Solutions
Problem 1 :
The angle of elevation of the top of the building at a distance of 50 m from its foot on a
horizontal plane is found to be 60 degree. Find the height of the building.
Solution :
Draw a sketch.
Here AB represents height of the building BC represents distance of the building from the point of
observation.
In the right triangle ABC the side which is opposite to the angle 60 degree is known as
opposite side AB the side which is opposite to 90 degree is called hypotenuse side AC
and the remaining side is called adjacent side BC.

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Now we need to find the length of the side AB.
tanθ Opposite side/Adjacent side
tan 60° AB/BC
√3 AB/50
√3 x 50 AB
AB 50√3
Approximate value of √3 is 1.732
AB 50 1.732
AB 86.6 m
So the height of the building is 86.6 m.
Problem 2 :
A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the
ladder is inclined at an angle of 60 degree. Find how far the ladder is from the foot of the
wall.
Solution :
Draw a sketch.
Here AB represents height of the wall BC represents the distance between the wall and the foot of
the ladder and AC represents the length of the ladder.

slide 3:

In the right triangle ABC the side which is opposite to angle 60 degree is known as opposite
side AB the side which is opposite to 90 degree is called hypotenuse side AC and
remaining side is called adjacent side BC.
Now we need to find the distance between foot of the ladder and the wall. That is we have to
find the length of BC.
tan θ Opposite side/Adjacent side
tan60° AB/BC
√3 6/BC
BC 6/√3
BC 6/√3 x √3/√3
BC 6√3/3
BC 2√3
Approximate value of √3 is 1.732
BC 2 1.732
BC 3.464 m
So the distance between foot of the ladder and the wall is 3.464 m.
Problem 3 :
A string of a kite is 100 meters long and it makes an angle of 60° with horizontal. Find the
height of the kiteassuming that there is no slack in the string.
Solution :
Draw a sketch.
Here AB represents height of kite from the ground BC represents the distance of kite from
the point of observation.

slide 4:

In the right triangle ABC the side which is opposite to angle 60 degree is known as opposite
side AB the side which is opposite to 90 degree is called hypotenuse side AC and
remaining side is called adjacent side BC.
Now we need to find the height of the side AB.
Sin θ Opposite side/Hypotenuse side
sinθ AB/AC
sin 60° AB/100
√3/2 AB/100
√3/2 x 100 AB
AB 50 √3 m
So the height of kite from the ground 50 √3 m.
Problem 4 :
From the top of the tower 30 m height a man is observing the base of a tree at an angle of
depression measuring 30 degree. Find the distance between the tree and the tower.
Solution :
Draw a sketch.
Here AB represents height of the tower BC represents the distance between foot of the tower
and the foot of the tree.
Now we need to find the distance between foot of the tower and the foot of the tree BC.
tan θ Opposite side/Adjacent side
tan 30° AB/BC
1/√3 30/BC
BC 30√3

slide 5:

Approximate value of √3 is 1.732
BC 30 1.732
BC 81.96 m
So the distance between the tree and the tower is 51.96 m.
Problem 5 :
A man wants to determine the height of a light house. He measured the angle at A and found
that tan A 3/4. What is the height of the light house if A is 40 m from the base
Solution :
Draw a sketch.
Here BC represents height of the light house AB represents the distance between the light
house from the point of observation.
In the right triangle ABC the side which is opposite to the angle A is known as opposite side
BC the side which is opposite to 90 degree is called hypotenuse side AC and remaining
side is called adjacent side AB.
Now we need to find the height of the light house BC.
tanA Opposite side/Adjacent side
tanA BC/AB
Given : tanA 3/4

slide 6:

3/4 BC/40
3 x 40 BC x 4
BC 3 x 40/4
BC 3 x 10
BC 30 m
So the height of the light house is 30 m.
Problem 6 :
A man wants to determine the height of a light house. He A ladder is leaning against a
vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the
wall.Find the length of ladder.
Solution :
Draw a sketch.
Here AB represents height of the wall BC represents the distance of the wall from the foot of
the ladder.
In the right triangle ABC the side which is opposite to the angle 20 degree is known as
opposite side ABthe side which is opposite to 90 degree is called hypotenuse side AC and
remaining side is called adjacent side BC.
Now we need to find the length of the ladder AC.
Cos θ Adjacent side/Hypotenuse side
Cos θ BC/AC
Cos 20° 3/AC

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0.9396 3/AC
AC 3/0.9396
AC 3.192
So the length of the ladder is 3.192 m.
Problem 7 :
A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What
is the length of string
Solution :
Draw a sketch.
Here AB represents height of the kite. In the right triangle ABC the side which is opposite to
angle 31 degree is known as opposite side AB the side which is opposite to 90 degree is
called hypotenuse side AC and the remaining side is called adjacent side BC.
Now we need to find the length of the string AC.
Sin θ Opposite side/Hypotenuse side
Sin θ AB/AC
Sin 31° AB/AC
0.5150 65/AC
AC 65/0.5150
AC 126.2 m
Hence the length of the string is 126.2 m.

slide 8:

Problem 8 :
The length of a string between a kite and a point on the ground is 90 m. If the string is
making an angle θ with the level ground such that tan θ 15/8 how high will the kite be
Solution :
Draw a sketch.
Here AB represents height of the balloon from the ground. In the right triangle ABC the side
which is opposite to angle θ is known as opposite side AB the side which is opposite to 90
degree is called hypotenuse side AC and remaining side is called adjacent side BC.
Now we need to find the length of the side AB.
Tan θ 15/8 ——– Cot θ 8/15
Csc θ √1+ cot²θ
Csc θ √1 + 64/225
Csc θ √225 + 64/225
Csc θ √289/225
Csc θ 17/15 ——- Sin θ 15/17
But sin θ Opposite side/Hypotenuse side AB/AC
AB/AC 15/17
AB/90 15/17
AB 15 x 90/17
AB 79.41
So the height of the tower is 79.41 m.

slide 9:

Problem 9 :
An aeroplane is observed to be approaching the airpoint. It is at a distance of 12 km from the
point of observation and makes an angle of elevation of 50 degree. Find the height above the
ground.
Solution :
Draw a sketch.
Here AB represents height of the airplane from the ground.In the right triangle ABC the side
which is opposite to angle 50 degree is known as opposite side AB the side which is
opposite to 90 degree is called hypotenuse side AC and remaining side is called adjacent
side BC.
Now we need to find the length of the side AB.
From the figure given above AB stands for the height of the aeroplane above the ground.
sin θ Opposite side/Hypotenuse side
sin 50° AB/AC
0.7660 h/12
0.7660 x 12 h
h 9.192 km
So the height of the aeroplane above the ground is 9.192 km.
Problem 10 :
A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60
degree angle . Find the height of the balloon from the ground. Imagine that there is no slack
in the cable

slide 10:

Solution :
Draw a sketch.
Here AB represents height of the balloon from the ground. In the right triangle ABC the side
which is opposite to angle 60 degree is known as opposite side AB the side which is
opposite to 90 degree is called hypotenuse AC and the remaining side is called as adjacent
side BC.
Now we need to find the length of the side AB.
From the figure given above AB stands for the height of the balloon above the ground.
sin θ Opposite side/Hypotenuse side
sin θ AB/AC
sin 60° AB/200
√3/2 AB/200
AB √3/2 x 200
AB 100√3
Approximate value of √3 is 1.732
AB 100 1.732
AB 173.2 m
So the height of the balloon from the ground is 173.2 m.
After having gone through the stuff given above we hope that the students would have
understood how to solve word problems in trigonometry.
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