Construct a triangle ΔABC with ∠B=50° AC=6 cm AB+BC=8 cm see also Q42942.


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Question Number 43135 by MrW3 last updated on 07/Sep/18

$C o n s t r u c t a t r i a n g l e Δ A B C w i t h$ $∠ B = 50 °$ $A C = 6 c m$ $A B + B C = 8 c m$ $s e e a l s o Q 42942 .$
Commented by math1967 last updated on 08/Sep/18

Commented by math1967 last updated on 08/Sep/18

$P C = A B + B C, ∠ A P B = ∠ P A B = 25 °$ $F r o m C, l o c a t e p t . A o n P X s u c h A C = 6 c m$ $F r o m A c o n s t r u c t ∠ P A B e q u a l ∠ A P B m e e t s$ $P C a t B . △ A B C i s r e q u i r e d t r i a n g l e$
Commented by MrW3 last updated on 08/Sep/18

$V e r y n i c e, t h a n k y o u s i r!$
Commented by math1967 last updated on 08/Sep/18

$Y o u a r e w e l c o m e s i r$
	Answered by behi83417@gmail.com last updated on 07/Sep/18
	$b=6,a+c=8 b^2 =a^2 +c^2 −2ac.cosB=(a+c)^2 −2ac(1+cosB) ⇒36=64−2ac(1+.64)⇒ac=8.54 ⇒ { ((a+c=8)),((ac=8.54)) :}⇒ { ((a=6.72,c=1.27)),((a=1.27,c=6.72)) :} now you can draw the triangle.$
	$b = 6, a + c = 8$ $b^{2} = a^{2} + c^{2} - 2 a c . c o s B = {(a + c)}^{2} - 2 a c (1 + c o s B)$ $\Rightarrow 36 = 64 - 2 a c (1 + . 64) \Rightarrow a c = 8.54$ $\Rightarrow {\begin{cases} a + c = 8 \\ a c = 8.54 \end{cases} \Rightarrow {\begin{cases} a = 6.72, c = 1.27 \\ a = 1.27, c = 6.72 \end{cases}$ $n o w y o u c a n d r a w t h e t r i a n g l e .$
	Commented by MrW3 last updated on 07/Sep/18

	$t h a n k y o u s i r .$ $b u t w e n e e d a w a y t o d r a w t h e t r i a n g l e$ $o n l y w i t h r u l e r a n d c o m p a s s w i t h o u t$ $c a l c u l a t i o n, l i k e i n Q 42942 .$
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