working out 'x' which is a side of a scalene inside a right-angle

**Jamie88** · April 10th 2008, 09:21 AM

Hello everybody.

I was just wondering how one would calculate the length of x on the diagram above. The questions after this in my work are easier so I think I must be missing something easy here as well.

it's not tan 50 100/x is it...?

Any help would be brill, thank you.

**Moo** · April 10th 2008, 09:26 AM

Hello,

If we name from up to down the points on the right side : A (summit), B, C, D (summit of right angle), what is x ? ;-)

**Jamie88** · April 10th 2008, 09:45 AM

That's confused me... :s

**Soroban** · April 10th 2008, 05:31 PM

Hello, Jamie!

[quote]

Code:

    A o
      * *
    x *50°*
      *     *
      *       *
    D o         *
      *   *       *
      *       *  10°*
      *           *   * 
      *          30°  * *
    C o * * * * * * * * * o B
              100

In right triangle $DCB\!:\;\;\tan30^o \:=\:\frac{CD}{100}\quad\Rightarrow\quad CD \:=\:100\tan30^o\;\;{\color{blue}[1]}$

In right triangle $ACB\!:\;\;\tan40^o \:=\:\frac{AC}{100}\quad\Rightarrow\quad AC \:=\:100\tan40^o\;\;{\color{blue}[2]}$

But $AC \:=\:x + CD\quad\Rightarrow\quad x \:=\:AC - CD\;\;{\color{blue}[3]}$

Substitute [1] and [2] into [3]: . $\boxed{x \;=\;100\tan40^o - 100\tan30^o}$

**Jamie88** · April 13th 2008, 07:29 AM

Brilliant, thank you for your help.

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