# Math Help - Compound Angles

1. ## Compound Angles

My answer is nearly right, except for the sign. Can anyone help me out here?

Many thanks.

Q.
The diagram (see attachment) shows a triangle of height h metres. The angles A & B are such that $A+B=45^{\circ}$. By using the expansion of $\tan (A+B)$, or otherwise, find the value of h.

Attempt: $\tan 45^{\circ}=1$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=1$
$\frac{(h/2)+(h/3)}{1-(h/2)(h/3)}=1$
$\frac{h}{2}+\frac{h}{3}=1-\frac{h^2}{6}$
$\frac{h^2}{6}+\frac{h}{2}+\frac{h}{3}-1=0$
$h^2+3h+2h-6=0$
$h^2+5h-6=0$
$(h-1)(h+6)=0$
h = 1 or -6

Ans.: (From text book): h = 6

Tan A= 2/h.

Thank you.