How Is Trigonometry Used to Calculate Flagpole Heights?

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The discussion focuses on calculating the height of a flagpole using trigonometric principles, specifically the tangent function. Given a horizontal distance of 120 meters from the wall, the angles of elevation to the top and bottom of the flagpole are 30 degrees and 25 degrees, respectively. The calculations yield a height of approximately 13.33 meters for the flagpole. The importance of avoiding intermediary rounding during calculations is emphasized to ensure accuracy.

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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

- - - Updated - - -

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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