Question

1stangle(∠DBE)=(x+3)
∘

\sf{2nd\: angle \: (\angle{DBC}) = {(x + 20)}^\circ}2ndangle(∠DBC)=(x+20)
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\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
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=> \sf{x + {3}^\circ + x + {20}^\circ + x + {7}^\circ = {180}^\circ}x+3
∘
+x+20
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+x+7
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=180
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=> \sf{3x + {30}^\circ = {180}^\circ}3x+30
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=180
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=> \sf{3x = {180}^\circ - {30}^\circ}3x=180
∘
−30
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=> \sf{3x = {150}^\circ}3x=150
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=> \sf{x = \dfrac{150^\circ}{3}}x=
3
150
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​


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

The measure of each angle:-

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

Answer 1

Answer:

${\angle{CBF} = {(x + 7)}^\circ = {(50 + 7)}^\circ = {57}^\circ}∠CBF=(x+7) </p><p>∘</p><p>=(50+7) </p><p>∘</p><p>=57$

Answer 2

angle EBD=(x+3)

angle CBF=(x+7)

angle DBC=(x+20)

so,angle EBC=(x+17)

EBD+EBC+CBF=180

x+3+×+17+×+7=180

or 3x+27=180

or 3x=180-27

or3x=153

or x=51

Explanation:

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