Question

Question 7 (b)
(b) Find x+y+z+w

Answer 1

(b) Using angle sum property of a quadrilateral,
${60}^{\circ }+{80}^{\circ }+{120}^{\circ }+n={360}^{\circ }\Rightarrow {260}^{\circ }+n={360}^{\circ }\Rightarrow n={360}^{\circ }-{260}^{\circ }\Rightarrow n={100}^{\circ }$

Since sum of supplementary angles is ${180}^{\circ }$.
$\therefore w+100={180}^{\circ }...\left(i\right)x+{120}^{\circ }={180}^{\circ }...\left(ii\right)y+{80}^{\circ }={180}^{\circ }...\left(iii\right)z+{60}^{\circ }={180}^{\circ }...\left(iv\right)$
Adding eq. (i), (ii), (iii), and (iv),
$\Rightarrow x+y+z+w+{100}^{\circ }+{120}^{\circ }+{80}^{\circ }+{60}^{\circ }={180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ }$
$\Rightarrow x+y+z+w+{360}^{\circ }={720}^{\circ }$
$\Rightarrow x+y+z+w={720}^{\circ }-{360}^{\circ }$
$\Rightarrow x+y+z+w={360}^{\circ }$