# Trigonometry problem concerning compound angles

• Mar 5th 2012, 01:02 PM
mathshelp94
Trigonometry problem concerning compound angles
The diagram (attached below) shows a right-angled triangle ABC in which |AB| = hm

By using the formula for tan(theta + 45 deg), or otherwise, find the two possible values for h.

Thanks!

Diagram:
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• Mar 5th 2012, 08:24 PM
Soroban
Re: Trigonometry problem concerning compound angles
Hello, mathshelp94!

Quote:

Code:

      A       *       |**       |@* *       |  *45*     h |  *  *       |    *    *       |    *    *       |      *      *     B *-------*-------* C       : - 1 - D - 5 - :
By using the formula for tan(theta + 45 deg), or otherwise, find the two possible values for h.

$\displaystyle \text{We see that: }\:\begin{Bmatrix}\tan\theta \:=\:\dfrac{1}{h} \\ \tan(\theta + 45^o) \:=\:\dfrac{6}{h} \end{Bmatrix}$

$\displaystyle \tan(\theta + 45^o) \;=\;\dfrac{\tan\theta + \tan45^o}{1 - \tan\theta\tan45^o} \:=\:\dfrac{6}{h}$

. . . . . . . . . . . . . $\displaystyle \dfrac{\frac{1}{h} + 1}{1 - \frac{1}{h}} \;=\;\dfrac{6}{h}$

Multiply by $\displaystyle \frac{h}{h}\!:\quad\;\;\dfrac{1 + h}{h - 1} \;=\;\dfrac{6}{h} \quad\Rightarrow\quad h+h^2 \:=\:6h - 6$

. . $\displaystyle h^2 - 5h + 6 \:=\:0 \quad\Rightarrow\quad (h-2)(h-3) \:=\:0 \quad\Rightarrow\quad \boxed{h \:=\:2\text{ or }3}$