# Applications of Right Triangles

• Jun 2nd 2008, 11:35 AM
lax600
Applications of Right Triangles
24.) The angle of elevation of a flagpole from a point level with its base is 45 degrees. from a point 20 ft. further away, the angle of elevation is 30 degrees. What is the hight of the flagpole to the nearest 0.1 ft.?

I did:

x*tan(45) = (x + 20)*tan(30)
x*tan(45) = x*tan(30) + 20tan(30)
x*tan(45) - x*tan(30) = + 20tan(30)
x(tan45 - tan30) = 20tan(45)
x = 20tan(30)/tan(45)-tan
x = 27.29

or is it:

x*tan(30) = (x - 20)*tan(45)
x*tan(30) = x*tan(45) -20tan(45)
x*tan(30) - tan(45) = -20tan(45)
x(tan30 - tan45) = -20tan(45)
x = -20tan(45)/tan(30) - tan(45)
x = 47.28

I don't know which one is right....(Crying)
• Jun 2nd 2008, 12:53 PM
Soroban
Hello, lax600!

The problem is easier than you think ... Note that 45° angle!

Quote:

The angle of elevation of a flagpole from a point level with its base is 45°.
From a point 20 ft. further away, the angle of elevation is 30°.
What is the height of the flagpole to the nearest 0.1 ft.?

Code:

    A *       | * *       |  *  *       |    *    *     x |      *      *       |        *        *       |          *          *       |        45° *        30° *       * - - - - - - - * - - - - - - - *       B      x      C      20      D

Since $\angle ACB = 45^o,\;AB = BC = x$

In right triangle $ABD\!:\;\;\tan30^o \:=\:\frac{x}{x+20}$

Solve for $x\!:\;\;x \;=\;\frac{20\tan30^o}{1 - \tan30^o}$

Since $\tan30^o \,=\,\frac{1}{\sqrt{3}}$, we have: . $x \;=\;\frac{20\cdot\frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \;=\;10(\sqrt{3}+1) \;\approx\;27.3$ ft.

• Jun 2nd 2008, 01:21 PM
lax600