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Can You Please Help Me class=

Sagot :

Answer:

In ΔECD

<E = <C = <D

using angle sum property of triangle

<E + <C + <D = 180°

<E + <E + <E = 180° ( As <E = <C = <D )

3<E = 180

<E = 60°

NOW

<BCA + <ACE + <ECD = 180 (L. P. A.)

50° + <ACE + 60° =180

110 + <ACE = 180

<ACE = 70°

NOW as

AC = AE

So using property of triangle that angle opposite to equal sides are equal we get

<ACE = <AEC = 70°

In triangle ΔAEC

Now using Angle sum property of triangle

<ACE + <AEC + <CAE = 180°

70° + 70° + x = 180

140 + x = 180

x = 40°

Now in ΔABC

AB = AC

using property of triangle that angle opposite to equal sides are equal

So <ACB = <ABC = 50°

Now using angle sum property of triangle in ΔABC

<ACB + <ABC + <CAB = 180

50 + 50 + y = 180

100 + y = 180

y = 80°

so x = 40° and y = 80°

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