# Trigonometry - worked solutions

Trigonometric Ratios with solutions

O

A

O

cos θ =

tan θ =

H

H

A

where O = opposite, A = adjacent and H = hypotenuse.

sin θ =

1. The angle of elevation of the top of a tree from a point on the ground 10m from the base of the tree is 28◦ . What is the height of the tree?

Solution

We need to find the height of tree which is opposite an angle of 28◦ .Thus O

tan θ =

A

O = A tan θ

height of tree = 10 × tan 28◦

=⇒ height of tree = 5.3m 1d.p.

2. Using a surveying instrument 1.6m high, the angle of elevation of the top of a tower is measured to be 55◦ from a point 6m from the base of the tower. How high is the tower? Solution

The surveying instrument is already at a height of 1.6m. So we need to find the height of the tower using the instrument, then add this to 1.6m. O

A

O = A tan θ

height of tower using instrument = 6 × tan 55◦

=⇒ height of tower using instrument = 4.2m 1d.p.

tan θ =

The height of the tower is 4.2m+1.6m=10.2m high.

3. The angle of elevation of the top of a 20m high mast from a point at ground level is 34◦ . How far is the point from the foot of the mast?

Solution

Let y be the distance of the mast from the point on the ground O

tan θ =

A

20

◦

tan 34 =

y

20

y =

tan 34◦

y = 6 × tan 55◦

=⇒ y = 29.7m 1d.p.

4. An isosceles triangle has base 8cm and sloping sides both with length 10cm. What is the base angle of this triangle? Solution

Split the base of the isosceles triangle in half. We now have a right-angled triangle of base 4cm and hypotenuse 10cm. We can now find the angle θ as follows

A

H

4

cos θ =

10

4

= 66.4218... = 66◦ to the nearest degree.

10

cos θ =

θ = cos−1

5. A supporting cable of length 30 m is fastened to the top of a 20m high mast. What angle does the cable make with the ground? How far away from the foot of the mast is it anchored to the ground?

Solution

We can use Pythagoras’ Theorem but we shall use trigonometry here. We first find the angle θ which is made when we look at the top of the mast from the point on the ground where the cable is fastened. So

sin θ =

θ = sin−1

20

40

O

H

= 41.8103149◦ = 41.8◦ to 1dp.

Now let y be the distance from the point on the ground to the foot of the mast.

tan θ =

20

y

20

tan 41.8103149◦

y = 22.4m 1dp

y =

6. A right-angled triangle has sides 5,12,13. What is the size of the smallest angle in this triangle?

Solution

Here 13 is the length of the hypotenuse, since it is the longest length in this right-angled triangle.

First we shall use trigonometry to find one of the angles which we shall call theta.

sin θ =

θ = sin−1

12

13

12

13

= 67.38013505◦ = 67.4◦ to 1dp.

We could find the other angle by using angles in a triangle sum to 180◦ but I can also use trigonometry.

sin α =

5

13

5

13

θ = 22.61986495

θ = 22.6◦ to 1dp

θ = sin−1

7. What is the height (to 1 d.p.) of an isosceles triangle with base angle 65◦ and sloping sides with length 10cm? What is the length of the base of this triangle? Solution

If we split the isosceles triangle in two we have two right-angled triangles. If we just look at one of these right-angled triangles then it has a base angle of 65◦ and hypotenuse 10cm. We can use this information to find the height (h) of the triangle, and x the length of the right-angled triangle. O

H

h

sin 65 =

10

h = 10 sin 65◦ = 9.06307787

=⇒ h = 9.1cm to 1dp.

sin θ =

We now solve for the base length x of the right-angled triangle. A

H

x

cos θ =

10

x = 10 cos 65◦ = 4.226182617

cos θ =

Therefore the base length of the isosceles triangle is 2x=2 × 4.226182617 =8.5cm to 1 decimal place.

8. One angle in a right angled triangle is 50◦ and the side opposite this angle has length 5cm. What is the length of the hypotenuse?

Solution

Here we simply need to find the hypotenuse H as follows

O

sin θ =

H

5

H =

= 6.5cm to 1dp

sin 50

9. In a right angled triangle, the hypotenuse has length 8m and one...

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