Question Video: Using Right-Angled Triangle Trigonometry to Solve Word Problems Involving Angles of Elevation | Nagwa Question Video: Using Right-Angled Triangle Trigonometry to Solve Word Problems Involving Angles of Elevation | Nagwa

# Question Video: Using Right-Angled Triangle Trigonometry to Solve Word Problems Involving Angles of Elevation Mathematics • First Year of Secondary School

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A mountain is 8.78 km tall from the ground. The angle of elevation of the top of the mountain from a point on the ground is 53°. Find the distance between the point on the ground and the top of the mountain, giving the answer to the nearest metre.

04:07

### Video Transcript

A mountain is 8.78 kilometres tall from the ground. The angle of elevation of the top of the mountain from a point on the ground is 53 degrees. Find the distance between the point on the ground and the top of the mountain, giving the answer to the nearest metre.

It’s always good to begin a question like this with a diagram. So we have our mountain first of all, which is 8.78 kilometres tall. We’re then told that the angle of elevation of the top of the mountain from a point on the ground is 53 degrees.

Now an angle of elevation is an angle measured from the horizontal to the line of sight when we look up towards an object. So on our diagram, that’s this angle here. We’re then asked to find the distance between the point on the ground and the top of the mountain, thus this distance here, which I’ve marked as 𝑑.

We can see that these three lines — so that’s the vertical height of the mountain, the horizontal ground, and the distance between the point on the ground and the top of the mountain — form a right-angled triangle. We know the length of one side in this triangle, the size of one of the other angles. And we’re looking to calculate the length of a second side, which means we can apply right angle trigonometry to this problem.

We begin by labeling the three sides of this triangle. The longest side of a right-angled triangle, which is the side opposite the right angle, is always the hypotenuse. The side diagonally opposite the other known angle — so that’s the angle of 53 degrees — is called the opposite. And the side between the known angle and the right angle is called the adjacent.

To help us decide which trigonometric ratio we need in this question, we can recall the acronym SOHCAHTOA, where S, C, and T stand for sin, cos, and tan and O, A, and H stand for opposite, adjacent, and hypotenuse. The side we know is the opposite, and the side we want to calculate is the hypotenuse. So we’re going to be using SOH, which is the sine ratio. The definition of the sine ratio is that sin of an angle 𝜃 is equal to the length of the opposite side divided by the length of the hypotenuse.

Before we start substituting into this formula though, notice that we’ve been asked to give our answer to the nearest metre, whereas the length we’ve been given for the opposite is in kilometres. We need to convert this measurement first. There are 1000 metres in a kilometre. So multiplying by 1000, we see that 8.78 kilometres is equivalent to 8780 metres.

Now we can substitute into the sine ratio. And we have that sin of 53 degrees is equal to 8780 over 𝑑. To solve this equation for 𝑑, we must first bring it out of the denominator on the right, which we do by multiplying both sides of the equation by 𝑑, giving 𝑑 sin 53 degrees is equal to 8780.

Next, we need to divide both sides of the equation by sin of 53 degrees, which we can do with no problem as sin of 53 degrees is just a number. At this point, we can use our calculator to evaluate this. But we must make sure that our calculator is in degree mode first. 8780 divided by sin of 53 degrees is 10993.75108.

Remember, we need to give our answer to the nearest metre. And as the deciding digit — so in this case, that’s the first number after the decimal point — is a seven, we round up. We have then the distance between the point on the ground and the top of the mountain to the nearest metre is 10994 metres.

Remember, we converted that measurement of 8.78 kilometres into metres before we did our trigonometry. However, we could’ve done trigonometry using that measurement in kilometres and then converted our answer into metres at the end before rounding by multiplying by 1000.

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