Question

find the answer for the question​

Answer 1

$\huge{\pink{\underline{\underline{\bf{☆\;SoluTion:}}}}}$

Given,

∆ABC ≈ ∆DEF

Such That,

AB = DE

BC = EF

CA = FD

Hence, By CPCT

☞ Angle ABC = Angle DEC

$→\;{\sf{(x+5)° = 50°}}$

$→\;{\sf{x = 50°-5°}}$

$→\;{\sf{x = 45°}}$

✒️ The Value of x is $\red{\bf{45°}}$

☞ Angle ACB = Angle DFE

$→\;{\sf{25° = (y-5)°}}$

$→\;{\sf{25°+5° = y}}$

$→\;{\sf{y = 30°}}$

✒️ The Value of y is $\green{\bf{30°}}$

Hope It Helps You ✌️

Answer 2

Answer:

Finding Angle B

Angle B = Angle E

So, x +5° = 50°

x = 50°- 5°

x = 45°

Finding Angle F

Angle F = Angle C

So, y - 5° = 25°

y = 25°+5°

y = 30°

Hope it will help you.

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