Height of a Tower

**purplec16** · February 28th 2010, 09:38 AM

From a point P on level ground, the angle of elevation of he top of the tower is 26.83 degrees. From a point 25.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53.5 degrees. Approximate the height of the tower.

**Jhevon** · February 28th 2010, 10:31 AM

Originally Posted by purplec16

From a point P on level ground, the angle of elevation of he top of the tower is 26.83 degrees. From a point 25.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53.5 degrees. Approximate the height of the tower.

See the attached diagram. Note that triangles PTB and ATB are both right triangles

Do you think you can figure out how to get to the answer?

**purplec16** · February 28th 2010, 11:02 AM

I know how to draw it I jus dont know how to work it

**Jhevon** · February 28th 2010, 11:17 AM

Originally Posted by purplec16

I know how to draw it I jus dont know how to work it

Use the big right triangle to solve for P (your answer will be in terms of h and require the use of trig functions). In a similar manner, use the small triangle to solve for P - 25. Your answer will also have h in it. Solve for P again there. Then, equate the two expressions you got for P and solve the resulting equation for h.

Think you can manage that? Try

**purplec16** · February 28th 2010, 12:18 PM

I tried, I didn't get it

**Jhevon** · February 28th 2010, 12:29 PM

Originally Posted by purplec16

I tried, I didn't get it

Post what you've tried so we can see where you're having trouble, and fix it.

**purplec16** · February 28th 2010, 12:39 PM

Ok u said to first solve for P so...
I did $sin 53.5= h/ P-25$
$25 sin 53.5 = h$
$P=45.09$

Then for the smaller triangle h would be
$20 sin 26.83 = 9.027$

You said something bout equate them, but the answer in the book is 20.2m, so

**Jhevon** · February 28th 2010, 12:54 PM

Originally Posted by purplec16

Ok u said to first solve for P so...
I did $sin 53.5= h/ P-25$
$25 sin 53.5 = h$
$P=45.09$

Then for the smaller triangle h would be
$20 sin 26.83 = 9.027$

You said something bout equate them, but the answer in the book is 20.2m, so

ok, first of all, sine does not come into play here. you have the opposite and adjacent sides!

secondly, how did you find out that P = 45.09 from the first triangle only?! you have one equation and two unknowns. There is no way you can solve for P like that! the answer has to have h in it.

make another attempt

**purplec16** · February 28th 2010, 01:13 PM

$tan 53.5 = h/25$

$25 tan 53.5= h$

$h= 33.79 m$

$tan 26.83 = h/P -25$

$25 tan 26.83= h$

$h= 12.64 m$

$P= 21.15$

So how do you find h?

**Jhevon** · February 28th 2010, 01:22 PM

Originally Posted by purplec16

$tan 53.5 = h/25$

$25 tan 53.5= h$

$h= 33.79 m$

how can you find h from what you have here? that 25 should not be there. the base of the small triangle is P - 25

$tan 26.83 = h/P -25$

$25 tan 26.83= h$

$h= 12.64 m$

$P= 21.15$

So how do you find h?

i do not know what happened here. first, your -25 in the first line just disappeared into thin air. then, you assume that P = 25, which it doesn't.

**purplec16** · February 28th 2010, 01:25 PM

Well I really dont understand what to do so can you just show me now please I tried twice

**Jhevon** · February 28th 2010, 01:37 PM

Originally Posted by purplec16

Well I really dont understand what to do so can you just show me now please I tried twice

From the big triangle,

$\tan 26.83 = \frac hP$

$\Rightarrow P = \frac h{\tan 26.83}$ ...............(1)

From the small triangle,

$\tan 53.5 = \frac h{P - 25}$

$\Rightarrow P = \frac h{\tan 53.5} + 25$ .............(2)

Now equate (1) and (2)

$\Rightarrow \frac h{\tan 26.83} = \frac h{\tan 53.5} + 25$

now solve for $h$

**Archie Meade** · February 28th 2010, 02:35 PM

hi purplec,

There are 3 sides to a right-angled triangle and 3 angles.
One angle is 90 degrees and the other two are acute.

Sin, Cos and Tan give the ratio of 2 sides with respect to one of the acute angles.

You need to locate the sides "opposite to" and "adjacent to" an angle.
The longest side is the hypotenuse.

$SinA=\frac{opposite}{hypotenuse}$

$CosA=\frac{adjacent}{hypotenuse}$

$TanA=\frac{opposite}{adjacent}$

You can use SOHCAHTOA to help remember until you master it.

The small triangle is inside the big one.

The height "h" is the opposite to both angles and P is on the adjacent.

We use Tan if we are dealing with opposite and adjacent.

Big Triangle

$Tan(26.83^o)=\frac{h}{P}$

multiply both by P

$[P]Tan(26.83^o)=\frac{hP}{P}=h$

divide by the Tan term for P

$\frac{PTan(26.83^o)}{Tan(26.83^o)}=P=\frac{h}{Tan( 26.83^o)}$

Small Triangle

$Tan(53.5)=\frac{h}{(P-25)}$

multiply both sides by (P-25)

$(P-25)Tan(53.5^o)=\frac{(P-25)h}{(P-25)}=h$

Multiply out the left

$PTan(53.5^o)-25Tan(53.5^o)=h$

Bring the P term and h together by subtracting h from both sides

$PTan(53.5^o)-25Tan(53.5^o)-h=h-h=0$

add the 25Tan term to both sides

$PTan(53.5^o)-25Tan(53.5^o)+25Tan(53.5^o)-h=25Tan(53.5^o)$

$PTan(53.5^0)-h=25Tan(53.5^o)$

Use your result for h from the big triangle

$P=\frac{h}{Tan(26.83^o)}$

Therefore

$\frac{hTan(53.5^o)}{Tan(26.83^o)}-h=25Tan(53.5^o)$

multiply both sides by $Tan(26.83^o)$

$hTan(53.5^o)-hTan(26.83^o)=25Tan(53.5^o)Tan(26.83^o)$

Write this using h as a factor

$h[Tan(53.5^o)-Tan(26.83^o)]=25Tan(53.5^o)Tan(26.83^o)$

Divide both sides by the factor of h

$h=\frac{25Tan(53.5^o)Tan(26.83^o)}{Tan(53.5^o)-Tan(26.83^o)}$

It's of course simplest to use your calculator to find the two values of Tan.

**purplec16** · February 28th 2010, 03:33 PM

But I dont see how you transition in the big triangle and little triangle all into one formula
Use your result for h from the big triangle

**Archie Meade** · February 28th 2010, 04:19 PM

Hi purplec,

The big triangle gave us

$P=\frac{h}{Tan(26.83^o)}$

This does not allow us to find h unless we know P.
Or... to find P, we need to know h.

Remember how simultaneous equations work?

If there are 2 things you do not know, then you need a 2nd clue.
For instance x+2=5, then you know x is 3 or you can work it out.
That's one clue needed to find one unknown.

But if we had x+y =5, we are stuck.
They could be 0 and 5, or 1 and 4, or 2 and 3, or -2 and 7 and so on.

When we have 2 unknowns, we need 2 clues.
In simultaneous equations, you may have just cancelled terms as follows.

x+y=5
x-y=1

add them... 2x+y-y=5+1=6 gets rid of y, so 2x=6, so x=3 and then y=2

Or.... we could write x=1+y from the 2nd equation,

then use this in the other

(1+y)+y=5, so 2y+1=5, so 2y=5-1=4, so y=2 and x=3

Do you see the idea ?
You may have covered these in class.

We use the same type of trick here.

H is the vertical height of both triangles.

The bases are different, one is P, the other is P-25

However we can use Tan(angle), using both angles given to write
out our two clues, since the sides we are dealing with involve "opposite"
and "adjacent".

We write P "in terms of h" using one triangle,
just as we did with the x and y values above.

Then we swop that into the 2nd equation, effectively getting rid of P
just as we got rid of x above to figure out what y is.

We got rid of P (wrote it in terms of h) because the question asked us to find h.

Study this idea, because it's an important maths technique.

You are only getting used to trigonometry,
so this problem contains quite a bit of calculations.

But it's important that you master the idea of cancelling one of 2 unknowns
if you only want one of them.

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