Double-angle Formulae

**GrigOrig99** · November 7th 2012, 11:13 AM

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

Many thanks.

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

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

**Soroban** · November 7th 2012, 07:02 PM

Hello, GrigOrig99!

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

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

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

Code:

                                      o P
                                  * * |
                              *2B *   |
                          *     * 45-B|
                      *       *       | h
                  *         *         |
              *           *           |
          *             *             |
    Q o * * * * * * * o * * * * * * * o R
              x       S       a

Note that $\angle QPR \:=\:2B + 45^o-B \:=\:45^o+B$

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

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

Subtract [1] - [2]:

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

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

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

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

**GrigOrig99** · November 8th 2012, 02:37 AM

Great. Thank you very much.

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