# Double-angle Formulae

1. Nov 7, 2012

### odolwa99

In the final proof, I have $\frac{h-h\tan B}{\tan B}$, rather than $\frac{h-h\tan B}{1+\tan B}$. Can anyone help me out?

Many thanks.

1. The problem statement, all variables and given/known data

In the triangle pqr (see attachment), $|\angle qrp|=90^{\circ}$ & |rp| = h. s is a point on [qr] such that $|\angle spq|=2B$ & $|\angle rps|=45^{\circ}-B$, $0^{\circ}<B<45^{\circ}$. Show that $|sr|=h \tan(45^{\circ}-B)$

2. Relevant equations

3. The attempt at a solution

$h\tan(45-B)=h(\frac{\tan45-\tan B}{1+ tan45\cdot\tan B})=\frac{h-h\tan B}{1+\tan B}$

Apply the sine rule: $\frac{\sin(45-B)}{|sr|}=\frac{\sin(180-(90+45-B))}{h}$
$h(\sin(45-B))=|sr|\sin(45+B)$
$h(\sin\cos B-\cos45\sin B)=|sr|(\sin45\cos B+\cos45\sin B)$
$h(\frac{\cos B}{\sqrt{2}}-frac{sin B}{\sqrt{2}})=|sr|(\frac{\cos B}{\sqrt{2}}+\frac{\sin B}{\sqrt{2}})$
$|sr|=(\frac{h\cos B-h\sin B}{\sqrt{2}})/(\frac{\cos B}{\sin B}{\sqrt{2}})$
$\frac{h\sqrt{2}(\cos B- \sin B)}{\sqrt{2}(\cos B+\sin B})$
$h(\frac{\sin B}{\cos B}+\frac{\cos B}{\sin B})$
$h(-\tan B+\frac{1}{\tan B})$
$\frac{h-h\tan B}{\tan B}$

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2. Nov 7, 2012

### LCKurtz

What am I missing? I see right triangle srp with$$\tan(45-B) = \frac {sr} h$$Just solve for $sr$.

3. Nov 7, 2012

### odolwa99

Looks like I was over-thinking this. Ok, thank you.

4. Nov 7, 2012

### SammyS

Staff Emeritus
$\displaystyle \frac{\cos B- \sin B}{\cos B+\sin B}\ne\frac{-\sin B}{\cos B}+\frac{\cos B}{\sin B}$

You can't break up a fraction in that manner!

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