Question

Heya

plz help me

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Answer 1

Given:-

• $\sf{1st\: angle \: (\angle{DBE}) = {(x + 3)}^\circ}$
• $\sf{2nd\: angle \: (\angle{DBC}) = {(x + 20)}^\circ}$
• $\sf{3rd\: angle\: (\angle{CBF}) = {(x + 7)}^\circ}$

To find:-

• $\sf{Value\: of\: x }$

Solution:-

$\sf{\angle{DBE} + \angle{DBC} + \angle{CBF} = {180}^\circ\:\:\:\:(Linear\: Pair)}$

= $\sf{{(x + 3)}^\circ + {(x + 20)}^\circ + {(x + 7)}^\circ = {180}^\circ}$

=> $\sf{x + {3}^\circ + x + {20}^\circ + x + {7}^\circ = {180}^\circ}$

=> $\sf{3x + {30}^\circ = {180}^\circ}$

=> $\sf{3x = {180}^\circ - {30}^\circ}$

=> $\sf{3x = {150}^\circ}$

=> $\sf{x = \dfrac{150^\circ}{3}}$

=> $\sf{x = {50}^\circ}$

$\sf{\therefore\: The\: value\: of \: x \: is\: {50}^\circ}$

Now,

The measure of each angle:-

$\sf{\angle{DBE} = {(x + 3)}^\circ = {(50 + 3)}^\circ = {53}^\circ}$

$\sf{\angle{DBC} = {(x + 20)}^\circ ={( 50 + 20)}^\circ = {70}^\circ}$

$\sf{\angle{CBF} = {(x + 7)}^\circ = {(50 + 7)}^\circ = {57}^\circ}$

Answer 2

Answer:

1stangle(∠DBE)=(x+3)

∘

\sf{2nd\: angle \: (\angle{DBC}) = {(x + 20)}^\circ}2ndangle(∠DBC)=(x+20)

∘

\sf{3rd\: angle\: (\angle{CBF}) = {(x + 7)}^\circ}3rdangle(∠CBF)=(x+7)

∘

To find:-

\sf{Value\: of\: x }Valueofx

Solution:-

\sf{\angle{DBE} + \angle{DBC} + \angle{CBF} = {180}^\circ\:\:\:\:(Linear\: Pair)}∠DBE+∠DBC+∠CBF=180

∘

(LinearPair)

= \sf{{(x + 3)}^\circ + {(x + 20)}^\circ + {(x + 7)}^\circ = {180}^\circ}(x+3)

∘

+(x+20)

∘

+(x+7)

∘

=180

∘

=> \sf{x + {3}^\circ + x + {20}^\circ + x + {7}^\circ = {180}^\circ}x+3

∘

+x+20

∘

+x+7

∘

=180

∘

=> \sf{3x + {30}^\circ = {180}^\circ}3x+30

∘

=180

∘

=> \sf{3x = {180}^\circ - {30}^\circ}3x=180

∘

−30

∘

=> \sf{3x = {150}^\circ}3x=150

∘

=> \sf{x = \dfrac{150^\circ}{3}}x=

3

150

∘

=> \sf{x = {50}^\circ}x=50

∘

\sf{\therefore\: The\: value\: of \: x \: is\: {50}^\circ}∴Thevalueofxis50

∘

Now,

The measure of each angle:-

\sf{\angle{DBE} = {(x + 3)}^\circ = {(50 + 3)}^\circ = {53}^\circ}∠DBE=(x+3)

∘

=(50+3)

∘

=53

∘

\sf{\angle{DBC} = {(x + 20)}^\circ ={( 50 + 20)}^\circ = {70}^\circ}∠DBC=(x+20)

∘

=(50+20)

∘

=70

∘

\sf{\angle{CBF} = {(x + 7)}^\circ = {(50 + 7)}^\circ = {57}^\circ}∠CBF=(x+7)

∘

=(50+7)

∘

=57

∘

Step-by-step explanation:

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