((sin (2x+18°))/(sin (2x+12°))) =(√((sin 36°)/(sin 48°))) tan 2x = (√(tan M)) .(√(tan N)) 0°<M,N<90° ⇒M+N=?°


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Question Number 176468 by blackmamba last updated on 20/Sep/22

$\frac{\sin (2 x + 18 °)}{\sin (2 x + 12 °)} = \sqrt{\frac{\sin 36 °}{\sin 48 °}}$ $\tan 2 x = \sqrt{\tan M} . \sqrt{\tan N}$ $0 ° < M, N < 90 ° \Rightarrow M + N = ? °$
Commented byPeace last updated on 20/Sep/22

$i s t a r t e d W i t h e 48, i s e e n o n i c e F o r m$ $24 i s t h e R i g h t i t h i n k$
	Answered by Peace last updated on 20/Sep/22

	$\frac{s i n (2 x) c o s (18) + c o s (2 x) s i n (18)}{s i n (2 x) c o s (12) + c o s (2 x) s i n (12)} = \sqrt{\frac{s i n (2.18)}{s i n (2.12)}}$ $\Leftrightarrow \frac{t g (2 x) c o s (18) + s i n (18)}{t g (2 x) c o s (12) + \sin (12)} = \sqrt{\frac{\sin (36)}{\sin (24)} =}$ $t = t g (2 x)$ $\Leftrightarrow \frac{t g (2 x) + t g (18)}{t g (2 x) + t g (12)} = \frac{c o s (12)}{c o s (18)} . \sqrt{\frac{s i n (36)}{s i n (48)}} = \sqrt{\frac{s i n (36) {c o s}^{2} (12)}{{c o s}^{2} (18) s i n (24)}}$ $= \sqrt{\frac{2 s i n (18) c o s (18) {c o s}^{2} (12)}{{c o s}^{2} (18) . 2 s i n (24) c o s (24)}}$ $= \sqrt{\frac{t g (18) {c o s}^{2} (12)}{2 s i n (12) c o s (12)}} = \sqrt{\frac{t g (18)}{t g (12)}}$ $\frac{t + t g (18)}{t + t g (12)} = c \Rightarrow t (1 - c) = c t g (12) - t g (18) \Rightarrow t g (2 x) = \frac{c t g (12) - t g (18)}{c - 1}$ $\frac{\sqrt{t g (18) t g (12)} - t g (18)}{1 - \sqrt{\frac{t g (18)}{t g (12)}}} = \frac{\sqrt{t g (18)} (\sqrt{t g (12} - \sqrt{t g (18)})}{\sqrt{t g (12)} - \sqrt{t g (18)}} . \sqrt{t g (12)}$ $= \sqrt{t g (18) t g (12)}$ $M + N = 30$
	Commented byblackmamba last updated on 26/Sep/22

	$w r o n g$
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