MHB How Is Trigonometry Used to Calculate Flagpole Heights?

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To calculate the height of a flagpole fixed to a vertical wall, trigonometric functions are employed using angles of elevation from a point 120 meters away. The tangent of the angles (25 degrees to the bottom and 30 degrees to the top of the flagpole) is used to create equations that relate the height of the flagpole and the height of the wall. By eliminating variables through substitution, the height of the flagpole is determined to be approximately 13.33 meters. It is emphasized that maintaining precision in calculations is crucial, as rounding too early can lead to inaccuracies. Understanding the relationship between angles of elevation is also highlighted, noting that a higher angle indicates a greater height over the same horizontal distance.
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A vertical flagpole is fixed at the top of a vertical wall. From a point which is 120m measured horizontally from the base of the wall the angle of elevation to the top of the flagpole is 30 degrees, and the angle of elevation to the bottom of the flagpole is 25 degrees.

a) Draw a clearly labelled diagram to represent this situation

b) Calculate the length of the flagpole correct to 2 dp

View attachment 3811

I have attempted to draw the diagram, and hopefully figured the height out. Although I could be completely wrong!
 

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I would draw the diagram as follows:

View attachment 3812

From this, we see:

$$\tan\left(25^{\circ}\right)=\frac{w}{120}\tag{1}$$

$$\tan\left(30^{\circ}\right)=\frac{w+h}{120}\tag{2}$$

Now, can you use (1) to eliminate $w$ in (2)?
 

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MarkFL said:
I would draw the diagram as follows:

View attachment 3812

From this, we see:

$$\tan\left(25^{\circ}\right)=\frac{w}{120}\tag{1}$$

$$\tan\left(30^{\circ}\right)=\frac{w+h}{120}\tag{2}$$

Now, can you use (1) to eliminate $w$ in (2)?

$$\tan\left(25^{\circ}\right)=\frac{w}{120}\tag{1}$$

$$w=tan(25)x120=55.95$$

$$tan(30)=\frac{55.95+h}{120}$$

$$\therefore tan(30) \cdot 120 = 55.95+h$$

$$69.28+55.95+h$$

$$69.28-55.95=h$$

$$\therefore h=13.33$$

I think that's right

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If I am given a similar question with two angles. The biggest angle would obviously be the outer angle?
 
I would rewrite (2) as:

$$\tan\left(30^{\circ}\right)=\frac{w}{120}+\frac{h}{120}$$

Then use (1) to obtain:

$$\tan\left(30^{\circ}\right)=\tan\left(25^{\circ}\right)+\frac{h}{120}$$

And then solving for $h$, we get (in meters):

$$h=120\left(\tan\left(30^{\circ}\right)-\tan\left(25^{\circ}\right)\right)\approx13.33$$

Your answer is correct, however, it is best to obtain the exact value first, and then only at the very end do your rounding. Sometimes intermediary rounding can cause your end result to be inaccurate.

Regarding which angle is larger, we know that if two rays have the same terminus, but one passes through a higher point over the same horizontal distance, then its angle of elevation must be greater. :D
 
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