1. ## Double-angle Formulae

Having a bit of trouble with this one. Can anyone help me out?

Many thanks.

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

Attempt: $\displaystyle \frac{|qr|}{h}=\tan(2B+45-B)$
$\displaystyle |qr|=h\tan(45+B)$
$\displaystyle |qr|-|sr|=h\tan(45+B)-h\tan(45-B)$
$\displaystyle |qs|=\frac{h\tan45+h\tan B}{1-\tan45\tan B}-(\frac{h\tan45-h\tan B}{1+\tan45\tan B})$
$\displaystyle |qs|=\frac{2h\tan B}{1-\tan^245\tan B}$

2. ## Re: Double-angle Formulae

Hello, GrigOrig99!

By the way, don't you know that $\displaystyle \tan45^o = 1\,?$

Having a bit of trouble with this one.
No wonder . . . the problem is stated incorrectly.

$\displaystyle \text{In }\Delta PQR,\,\angle QRP=90^o\,\text{ and }\,RP = h.\;S\text{ is a point on }QR$
$\displaystyle \text{ such that }\angle SPQ= 2B\text{ and }\angle RPS =45^o-B,\;0^o<B<45^o$
$\displaystyle \text{If }SR=a=h\tan(45-B),\,\text{ show that: }\:x=QS = 2h\tan ({\color{red}2}B)$.
Code:
                                      o P
* * |
*2B *   |
*     * 45-B|
*       *       | h
*         *         |
*           *           |
*             *             |
Q o * * * * * * * o * * * * * * * o R
x       S       a
Note that $\displaystyle \angle QPR \:=\:2B + 45^o-B \:=\:45^o+B$

In $\displaystyle \Delta QPR\!:\;\tan(45+B) \,=\,\frac{x+a}{h} \quad\Rightarrow\quad x+a \:=\:h\tan(45+B)$ .[1]

In $\displaystyle \Delta SPR\!:\;\tan(45-B) \,=\,\frac{a}{h} \quad\Rightarrow\quad a \:=\:h\tan(45-B)$ .[2]

Subtract [1] - [2]:

. . $\displaystyle x \;=\;h\tan(45+b) - h\tan(45-B) \;=\;h\left[\tan(45+B) - \tan(45-B)\right]$

. . $\displaystyle x \;=\;h\left[\frac{\tan45 + \tan B}{1-\tan45\tan B} - \frac{\tan45 - \tan B}{1 + \tan45\tan B}\right] \;=\;h\left[\frac{1+ \tan B}{1-\tan B} - \frac{1 - \tan B}{1+\tan B}\right]$

. . $\displaystyle x \;=\;h\left[\frac{(1+\tan B)^2 - (1-\tan B)^2}{(1-\tan B)(1+\tan B)}\right] \;=\;h\left[\frac{1 + 2\tan B + \tan^2\!B - 1 + 2\tan B -\tan^2\!B}{1-\tan^2\!B}\right]$

. . $\displaystyle x \;=\;h\left[\frac{\4\tan B}{1-\tan^2\!B}\right] \;=\;2h\left[\frac{2\tan B}{1-\tan^2\!B}\right] \;=\;2h\tan(2B)$

3. ## Re: Double-angle Formulae

Great. Thank you very much.