# Compound Angles

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• Oct 31st 2012, 12:58 PM
GrigOrig99
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
• Oct 31st 2012, 01:04 PM
a tutor
Re: Compound Angles
Tan A= 2/h.
• Oct 31st 2012, 01:35 PM
GrigOrig99
Re: Compound Angles
Thank you.