Question

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Answer 1

in ∆tqr
<t+<q+<r=180°
90+40+x=180°
130+x=180°
x=180°-130°
x=50°

let the cross section point be o
in∆top
<p+<t+<o=180°
30+90+<o=180°
120+<o=180°
<o=180°-120°
<o=60°

in ∆qos
<q+<o+<s=180°
40+60+y=180°
100+y=180°
y=180°-100°
y=80°



x=50° and y=80°

Answer 2

in tringle qtr

<q+<t+<r[x]=180          <sum property

40+90+x=180

130+x=180

x=180-130     x=50deg

now in triangle pqs and triangle psr

<spq=<spr

qo=ot

ps=ps

therefore triangles are s≅    by s a s

qs=sr   by [cpct]

qr is a stright line

which is equal to 180

therefore

<y=90

hence showed