Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 187893 by Rupesh123 last updated on 23/Feb/23

	Answered by a.lgnaoui last updated on 23/Feb/23
	$△ABD ((sin 53)/(AD))=((sin (14+x))/(AB))=((sin (14+53+x))/(2k)) 2ksin 53=ADsin (67+x) (1) △ACD ((sin 14)/k)=((sin x)/(AD)) ksin x=ADsin 14 (2) (((1))/((2)))⇒ ((2sin 53)/(sin x))=((sin (67+x))/(sin 14)) 2sin 53×sin 14=sin x×sin (67+x) =sin^2 x×cos 67+sin xcos xsin 67 =cos 67sin^2 x+sin 67sin x(√(1−sin^2 x )) 1−sin^2 x=(2sin 53sin 14−cos 67sin^2 x)^2 ×(1/(sin^2 67sin^2 x)) posins sin x=z sin 53=a sin 14=b cos 67=c sin 67=d d^2 z^2 (1−z^2 )=(2ab−cz^2 )^2 d^2 z^4 −2d^2 z^2 +d^2 z^2 =4a^2 b^2 +c^2 z^4 −4abcz^2 z^2 =t (d^2 −c^2 )t^2 +(4abc−d^2 )t−4a^2 b^2 =0 t^2 +((4abc−d^2 )/(d^2 −c^2 ))t−((4a^2 b^2 )/(d^2 −c^2 ))=0 (t+((4abc−d^2 )/(2(d^2 −c^2 ))))^2 −(((4a^2 b^2 (d^2 −c^2 ))/(4(d^2 −c^2 )^2 ))+(((4abc−d^2 ))/(4(d^2 −c^2 )^2 )))=0 { ((t+((4abc−d^2 )/(2(d^2 −c^2 )))±((√(4a^2 b^2 (d^2 −c^2 )+4abc−d^2 ))/(2(d^2 −c^2 )))=0)),((t=sin^2 x x=arcsin (√(t.)) )) :} (a suivre)........$
	$△ A B D \frac{\sin 53}{A D} = \frac{\sin (14 + x)}{A B} = \frac{\sin (14 + 53 + x)}{2 k}$ $2 k \sin 53 = A D \sin (67 + x) (1)$ $△ A C D \frac{\sin 14}{k} = \frac{\sin x}{A D}$ $k \sin x = A D \sin 14 (2)$ $\frac{(1)}{(2)} \Rightarrow \frac{2 \sin 53}{\sin x} = \frac{\sin (67 + x)}{\sin 14}$ $2 \sin 53 \times \sin 14 = \sin x \times \sin (67 + x)$ $= \sin^{2} x \times \cos 67 + \sin x \cos x \sin 67$ $= \cos 67 \sin^{2} x + \sin 67 \sin x \sqrt{1 - \sin^{2} x}$ $1 - \sin^{2} x = {(2 \sin 53 \sin 14 - \cos 67 \sin^{2} x)}^{2} \times \frac{1}{\sin^{2} 67 \sin^{2} x}$ $p o s i n s \sin x = z$ $\sin 53 = a \sin 14 = b \cos 67 = c$ $\sin 67 = d$ $d^{2} z^{2} (1 - z^{2}) = {(2 a b - {c z}^{2})}^{2}$ $d^{2} z^{4} - 2 d^{2} z^{2} + d^{2} z^{2} = 4 a^{2} b^{2} + c^{2} z^{4} - 4 {a b c z}^{2}$ $z^{2} = t$ $(d^{2} - c^{2}) t^{2} + (4 a b c - d^{2}) t - 4 a^{2} b^{2} = 0$ $t^{2} + \frac{4 a b c - d^{2}}{d^{2} - c^{2}} t - \frac{4 a^{2} b^{2}}{d^{2} - c^{2}} = 0$ ${(t + \frac{4 a b c - d^{2}}{2 (d^{2} - c^{2})})}^{2} - (\frac{4 a^{2} b^{2} (d^{2} - c^{2})}{4 {(d^{2} - c^{2})}^{2}} + \frac{(4 a b c - d^{2})}{4 {(d^{2} - c^{2})}^{2}}) = 0$ ${\begin{cases} t + \frac{4 a b c - d^{2}}{2 (d^{2} - c^{2})} \pm \frac{\sqrt{4 a^{2} b^{2} (d^{2} - c^{2}) + 4 a b c - d^{2}}}{2 (d^{2} - c^{2})} = 0 \\ t = \sin^{2} x x = a r c \sin \sqrt{t .} \end{cases}$ $(a s u i v r e) . . . . . . . .$
	Answered by mr W last updated on 23/Feb/23

	$m = ‘ ‘ {m e d i a n}^{″}$ $\frac{m}{k} = \frac{\sin x}{\sin 14 °} . . . (i)$ $\frac{2 k}{m} = \frac{\sin (53 ° + 14 ° + x)}{\sin 53 °} . . . (i i)$ $(i) \times (i i) :$ $2 = \frac{\sin x}{\sin 14 °} \times \frac{\sin (53 ° + 14 ° + x)}{\sin 53 °}$ $2 \sin 14 ° \sin 53 ° = \sin^{2} x \cos 67 ° + \sin x \cos x \sin 67 °$ $4 \sin 14 ° \sin 53 ° = (1 - \cos 2 x) \cos 67 ° + \sin 2 x \sin 67 °$ $\cos 67 ° - 4 \sin 14 ° \sin 53 ° = \cos (2 x + 67 °)$ $x = \frac{\cos^{- 1} (\cos 67 ° - 4 \sin 14 ° \sin 53 °) - 67 °}{2}$ $\approx 22.732 °$
	Commented by Rupesh123 last updated on 23/Feb/23
	Very nice solution!
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum