MHB Angle of elevation and depression problems

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The discussion focuses on two problems involving angles of elevation and depression. In the first scenario, Professor Michael measures the height of a clock tower using angles of elevation of 30 and 40 degrees from two different distances, leading to a formula for calculating the height. The second problem involves an engineer determining the length of a tunnel between two points, using known distances and an angle to apply the cosine rule for calculation. Both scenarios illustrate practical applications of trigonometry in real-world contexts. The solutions involve setting up equations based on the tangent and cosine functions to find the required measurements.
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(1) Prof micheal walks towards the university clock tower and decides to find the height of the clock above ground. He determines the angle of elevation to be 30 degrees and after proceeding an additional 60m towards the base of the tower, he finds the angle of elevation to be 40 degrees. What is the height of the clock tower

(2) An engineer who is to dig a tunnel through a small mountain wish to deternine the length of the tunnel. Point X and Y are chosen as the end points of the tunnel, then a point Z is chosen from which the distances to X and Y are found to be 19km and 23km respectively. If angle XZY measures 48 degrees, find the length of the tunnel.
 
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If you draw out 1, you'll see two right angle triangles, one with an angle of 30 degrees with a distance of x metres from the clock tower, the other with an angle of 40 degrees with a distance of (x - 60) metres from the clock tower. Obviously both have a height of h metres, the clock tower's height. From there:

$\displaystyle \begin{align*} \tan{ \left( 30^{\circ} \right) } &= \frac{h}{x} \\ \frac{1}{\sqrt{3}} &= \frac{h}{x} \\ \frac{x}{\sqrt{3}} &= h \\ x &= h\sqrt{3} \end{align*}$

and also

$\displaystyle \begin{align*} \tan{ \left( 40^{\circ} \right) } &= \frac{h}{x - 60} \\ \tan{ \left( 40^{\circ} \right) } &= \frac{h}{h\sqrt{3} - 60} \\ \left( h\sqrt{3} - 60 \right) \tan{ \left( 40^{\circ} \right) } &= h \\ h\sqrt{3} \tan{ \left( 40^{\circ} \right) } - 60\tan{ \left( 40^{\circ} \right) } &= h \\ h\sqrt{3} \tan{ \left( 40^{\circ} \right) } - h &= 60\tan{ \left( 40^{\circ} \right) } \\ h \left[ \sqrt{3} \tan{ \left( 40^{\circ} \right) } - 1 \right] &= 60\tan{ \left( 40^{\circ} \right) } \\ h &= \frac{60\tan{ \left( 40^{\circ} \right) } }{\sqrt{3}\tan{ \left( 40^{\circ} \right) } - 1} \end{align*}$

Now put this into your calculator to get your decimal approximation :)

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Jerome said:
(1) Prof micheal walks towards the university clock tower and decides to find the height of the clock above ground. He determines the angle of elevation to be 30 degrees and after proceeding an additional 60m towards the base of the tower, he finds the angle of elevation to be 40 degrees. What is the height of the clock tower

(2) An engineer who is to dig a tunnel through a small mountain wish to deternine the length of the tunnel. Point X and Y are chosen as the end points of the tunnel, then a point Z is chosen from which the distances to X and Y are found to be 19km and 23km respectively. If angle XZY measures 48 degrees, find the length of the tunnel.

For the second, draw it out and you'll see you have a triangle with two known lengths and the angle between them known. You can use the cosine rule to evaluate the third length (which is the distance XY, the length of the tunnel).
 
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