Video: CBSE Class X • Pack 3 • 2016 • Question 26

CBSE Class X • Pack 3 • 2016 • Question 26

05:07

Video Transcript

The angle of elevation of the top 𝑄 of a vertical tower 𝑃𝑄 from a point 𝑋 on the ground is 60 degrees. From a point 𝑌 40 metres vertically above 𝑋, the angle of elevation of the top of the tower 𝑄 is 45 degrees. Find the height of the tower and the distance 𝑃𝑋. Use root three is equal to 1.73.

Our first step is to draw a diagram. The angle of elevation of 𝑄 from 𝑋 is 60 degrees. We know that 𝑃𝑄 is vertical. Therefore, 𝑃𝑄 is at right angles to 𝑃𝑋. 40 metres above point 𝑋 is point 𝑌. And we know that 𝑄 is at an angle of elevation of 45 degrees from 𝑌. We can therefore create a second right-angled triangle 𝑄𝑀𝑌.

We have been asked to calculate the height of the tower from 𝑃 to 𝑄, labelled ℎ, and the distance from 𝑃 to 𝑋, labelled 𝑏. In order to solve this problem, we need to use the trigonometrical ratios, often referred to as SOHCAHTOA.

In a right-angled triangle, the sin of angle 𝜃 is equal to the opposite over the hypotenuse. The cos or cosine of angle 𝜃 is equal to the adjacent divided by the hypotenuse. And the tan or tangent of angle 𝜃 is equal to the opposite divided by the adjacent.

If we firstly consider triangle 𝑄𝑀𝑌, then the tan of 45 degrees is equal to 𝑄𝑀 divided by 𝑀𝑌, as 𝑄𝑀 is opposite the 45-degree angle and 𝑀𝑌 is adjacent to the 45-degree angle. Tan of 45 degrees is equal to one. The length of 𝑄𝑀 is equal to ℎ minus 40, as the length of 𝑄𝑃 is equal to ℎ and the length of 𝑀𝑃 is equal to 40. The length of 𝑀𝑌 is equal to the length of 𝑃𝑋, in our diagram labelled 𝑏. Multiplying both sides of this equation by 𝑏 gives us 𝑏 is equal to ℎ minus 40. Therefore, the length of 𝑃𝑋 is equal to the height of 𝑃𝑄 minus 40. We will call this equation one.

We now need to consider the triangle 𝑄𝑃𝑋. In this triangle, tan 60 is equal to 𝑄𝑃 divided by 𝑃𝑋, as 𝑄𝑃 is opposite the 60-degree angle and 𝑃𝑋 is adjacent to the 60-degree angle. Tan of 60 is equal to root three, and we were told in the question to use root three equal to two decimal places, 1.73. 𝑄𝑃 is equal to ℎ, the height of the tower, and 𝑃𝑋 on the diagram is labelled 𝑏. Multiplying both sides of this equation by 𝑏 gives us a second equation 1.73𝑏 is equal to ℎ.

We now have two simultaneous equations that we can solve algebraically. There are lots of ways of approaching these simultaneous equations, but we will substitute equation two into equation one. Substituting ℎ is equal to 1.73𝑏 into equation one gives us 𝑏 is equal to 1.73𝑏 minus 40. Subtracting 𝑏 or one 𝑏 from both sides of this equation gives us zero is equal to 0.73𝑏 minus 40. Adding 40 to both sides of this equation gives us 40 is equal to 0.73𝑏. Dividing both sides by 0.73 gives us a value of 𝑏 of 54.79 to two decimal places. This means that the distance 𝑃𝑋 is equal to 54.79 metres.

Substituting this value back into equation two gives us a value of ℎ of 1.73 multiplied by 54.79. This is equal to 94.79 to two decimal places. Therefore, the height of the tower from 𝑃 to 𝑄 is 94.79 metres.

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