Trig Question-Please help!!

[aakanksha.sharma]

How exactly would I figure this out?

Angle measurements were taken from two points on direct opposite sides of a tree. One angle of elevation is 35 degrees and the other is 30 degrees. The distance between these two viewing points is 65m. How tall is the tree?

If you can help me it would be greatly appreciated, thanks!

 

[soroban]

Hello, aakanksha!

First, make a sketch . . .

Angle measurements were taken from two points on direct opposite sides of a tree.
One angle of elevation is 35 degrees and the other is 30 degrees.
The distance between these two viewing points is 65m.
How tall is the tree?

                  A
                  *
                * |  *
              *   |     *
            *     |h       *
          *       |           *
        * 35      |           30 *
      *  *  *  *  *  *  *  *  *  *  *
      C     x     B      65-x       D
      : - - - - - -  65 - - - - - - :

The height of the tree is \(\displaystyle h = AB.\)
The observation points are at \(\displaystyle C\) and \(\displaystyle D.\)
\(\displaystyle \angle C = 35^o,\;\angle D = 30^o,\;CD = 65\)
Let \(\displaystyle CB = x\), then \(\displaystyle BD \,=\,65-x\)

In right triangle \(\displaystyle ABC:\;\tan35 \,=\,\dfrac{h}{x} \quad\Rightarrow\quad x \:=\:\dfrac{h}{\tan35} \;\;\color{blue}{[1]}\)

In right triangle \(\displaystyle ABD:\;\tan30 \:=\:\dfrac{h}{65-x} \quad\Rightarrow\quad x \:=\: 65 - \dfrac{h}{\tan30}\;\;\color{blue}{[2]}\)


Equate [1] and [2]: .\(\displaystyle \dfrac{h}{\tan35} \:=\:65 - \dfrac{h}{\tan30}\)


Multiply by \(\displaystyle \tan30\tan35:\;h\tan30 \:=\:65\tan30\tan35 - h\tan35\)

. . . . . . . . . . . .\(\displaystyle h\tan30 + h\tan35 \:=\:65\tan30\tan35\)

. . . . . . . . . . . .\(\displaystyle h(\tan30 + \tan35) \:=\:65\tan30\tan35\)

. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle h \:=\:\dfrac{65\tan30\tan35}{\tan30 + \tan35}\)


Therefore: .\(\displaystyle h \;=\;20.56832618\)