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Question Number 183281 by mr W last updated on 24/Dec/22

Commented by mr W last updated on 24/Dec/22

$f i n d t h e l e n g t h o f A C .$
	Answered by mr W last updated on 25/Dec/22
	$cos α=((a^2 +q^2 −d^2 )/(2aq)) ⇒sin α=((√([q^2 −(a−d)^2 ][(a+d)^2 −q^2 ]))/(2aq)) cos β=((b^2 +q^2 −c^2 )/(2bq)) ⇒sin β=((√([q^2 −(b−c)^2 ][(b+c)^2 −q^2 ]))/(2bq)) p^2 =a^2 +b^2 −2ab cos (α+β) p^2 =a^2 +b^2 −2ab{((a^2 +q^2 −d^2 )/(2aq))×((b^2 +q^2 −c^2 )/(2bq))−((√([q^2 −(a−d)^2 ][(a+d)^2 −q^2 ]))/(2aq))×((√([q^2 −(b−c)^2 ][(b+c)^2 −q^2 ]))/(2bq))} p^2 =a^2 +b^2 −(((a^2 −d^2 +q^2 )(b^2 −c^2 +q^2 )−(√([(a+d)^2 −q^2 ][(a−d)^2 −q^2 ][(b+c)^2 −q^2 ][(b−c)^2 −q^2 ])))/(2q^2 )) p^2 =(((a^2 +b^2 +c^2 +d^2 −q^2 )q^2 −(a^2 −d^2 )(b^2 −c^2 )+(√([(a+d)^2 −q^2 ][(a−d)^2 −q^2 ][(b+c)^2 −q^2 ][(b−c)^2 −q^2 ])))/(2q^2 )) ⇒p=(1/q)(√(((a^2 +b^2 +c^2 +d^2 −q^2 )q^2 −(a^2 −d^2 )(b^2 −c^2 )+(√([(a+d)^2 −q^2 ][(a−d)^2 −q^2 ][(b+c)^2 −q^2 ][(b−c)^2 −q^2 ])))/2)) p=(1/(80))(√(((60^2 +78^2 +50^2 +100^2 −80^2 )×80^2 −(60^2 −100^2 )(78^2 −50^2 )+(√([(60+100)^2 −80^2 ][(60−100)^2 −80^2 ][(78+50)^2 −80^2 ][(78−50)^2 −80^2 ])))/2)) =30(√(17))≈123.693$
	$\cos α = \frac{a^{2} + q^{2} - d^{2}}{2 a q} \Rightarrow \sin α = \frac{\sqrt{[q^{2} - {(a - d)}^{2}] [{(a + d)}^{2} - q^{2}]}}{2 a q}$ $\cos β = \frac{b^{2} + q^{2} - c^{2}}{2 b q} \Rightarrow \sin β = \frac{\sqrt{[q^{2} - {(b - c)}^{2}] [{(b + c)}^{2} - q^{2}]}}{2 b q}$ $p^{2} = a^{2} + b^{2} - 2 a b \cos (α + β)$ $p^{2} = a^{2} + b^{2} - 2 a b {\frac{a^{2} + q^{2} - d^{2}}{2 a q} \times \frac{b^{2} + q^{2} - c^{2}}{2 b q} - \frac{\sqrt{[q^{2} - {(a - d)}^{2}] [{(a + d)}^{2} - q^{2}]}}{2 a q} \times \frac{\sqrt{[q^{2} - {(b - c)}^{2}] [{(b + c)}^{2} - q^{2}]}}{2 b q}}$ $p^{2} = a^{2} + b^{2} - \frac{(a^{2} - d^{2} + q^{2}) (b^{2} - c^{2} + q^{2}) - \sqrt{[{(a + d)}^{2} - q^{2}] [{(a - d)}^{2} - q^{2}] [{(b + c)}^{2} - q^{2}] [{(b - c)}^{2} - q^{2}]}}{2 q^{2}}$ $p^{2} = \frac{(a^{2} + b^{2} + c^{2} + d^{2} - q^{2}) q^{2} - (a^{2} - d^{2}) (b^{2} - c^{2}) + \sqrt{[{(a + d)}^{2} - q^{2}] [{(a - d)}^{2} - q^{2}] [{(b + c)}^{2} - q^{2}] [{(b - c)}^{2} - q^{2}]}}{2 q^{2}}$ $\Rightarrow p = \frac{1}{q} \sqrt{\frac{(a^{2} + b^{2} + c^{2} + d^{2} - q^{2}) q^{2} - (a^{2} - d^{2}) (b^{2} - c^{2}) + \sqrt{[{(a + d)}^{2} - q^{2}] [{(a - d)}^{2} - q^{2}] [{(b + c)}^{2} - q^{2}] [{(b - c)}^{2} - q^{2}]}}{2}}$ $p = \frac{1}{80} \sqrt{\frac{(60^{2} + 78^{2} + 50^{2} + 100^{2} - 80^{2}) \times 80^{2} - (60^{2} - 100^{2}) (78^{2} - 50^{2}) + \sqrt{[{(60 + 100)}^{2} - 80^{2}] [{(60 - 100)}^{2} - 80^{2}] [{(78 + 50)}^{2} - 80^{2}] [{(78 - 50)}^{2} - 80^{2}]}}{2}}$ $= 30 \sqrt{17} \approx 123.693$
	Commented by manxsol last updated on 24/Dec/22

	$b e h o l d t h e s u r p r i s e$
	Commented by manxsol last updated on 24/Dec/22

	Commented by mr W last updated on 25/Dec/22

	$i {c a n}^{'} t s e e w h a t y o u m e a n t w i t h t h e$ $s u r p r i s e . y o u s h o u l d s p e c i f y w h a t i s$ $g i v e n a n d w h a t i s t o f i n d .$
	Answered by Acem last updated on 24/Dec/22

	Commented by manxsol last updated on 25/Dec/22

	$i f w e p u t s i d e 78 u n k n o w n; w e h a v e t w o s o l u t i o n 123 a n d 50 . s a l u d o s S i r A c e m$
	Commented by Acem last updated on 25/Dec/22

	$S a l u d o s p a r a t i m i q u e r i d o a m i g o!$
	Commented by Acem last updated on 25/Dec/22

	$M y f r i e n d, d i d y o u m e a n t h a t t h e s i d e$ $B C i s u n k n o w n ?$ $i f n o t, t h e n h o w w i l l w e h a v e a n o t h e r s o l u t i o n$ $s u c h A C = 50 ?$
	Commented by Acem last updated on 25/Dec/22

	$D o y o u m e a n B C = 50 ? h o w ? i s a w y o u r s o l u t i o n$ $b u t B C i s 78 a n d t h e q u e s t i o n i s t o f i n d A C$ $W e l l, l e t m e k n o w w h a t d o y o u m e a n$
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