Question

please answer the question
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Answer 1

Answer:

ANGLE AOD+ANGLE DOC+ANGLE COB=180(ANGLES IN LINEAR PAIR)

x+10+x+x+20=180

3x+30=180

3x= 180-30=150

x=150/3

x=50

ANGLE COD= 50°

ANGLE AOD = x+10

= x+10=50+10=60

angle BOC= x+20=50+20= 70

Step-by-step explanation:

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Answer 2

★ How to do :-

Here, we are given with a diagram that has a straight line. This straight line is divided into three different parts. There are three angles formed in it. We aren't given with the correct measurement of each of the angle. But, we are given the value of each angles in teh form of variables and constants. We are asked to find the value of the variable x and the measurement of each of the angle in it. Here, we use some other concepts such as transition method, in which we shift the numbers and variables from one hand side to the other, which changes it's sign. So, let's solve!!

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➤ Solution :-

${\sf \longrightarrow \underline{\boxed{\sf Straight \: line \: angle \: measures \: {180}^{\circ}}}}$

Substitute the value of all angles.

${\tt \leadsto (x + 10)^{\circ} + {x}^{\circ} + (x + 20)^{\circ} = {180}^{\circ}}$

Add all the variables and constants separately.

${\tt \leadsto 3x + 30 = {180}^{\circ}}$

Shift the number 30 from LHS to RHS, changing it's sign.

${\tt \leadsto 3x = 180 - 30}$

Subtract the values on RHS.

${\tt \leadsto 3x = 150}$

Shift the number 3 from LHS to RHS.

${\tt \leadsto x = \dfrac{150}{3}}$

Simplify the fraction to get the value of x.

${\tt \leadsto x = {50}^{\circ}}$

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Now, let's find the value of each angles separately.

Value of ∠AOD :-

${\tt \leadsto x + 10 = 50 + 10}$

${\tt \leadsto \angle{AOD} = {60}^{\circ}}$

Value of ∠DOC :-

${\tt \leadsto \angle{DOC} = {50}^{\circ}}$

Value of ∠BOC :-

${\tt \leadsto x + 20 = 50 + 20}$

${\tt \leadsto \angle{COB} = {70}^{\circ}}$

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${\red{\underline{\boxed{\bf So, \: the \: value \: of \: x \: is \: 50.}}}}$

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$\begin{gathered} \small \boxed{\begin{array} {cc} \large \dag \: \sf Answers \\ \\ \sf \bigstar \: \angle{AOD} = {60}^{\circ} \\ \\ \sf \bigstar \: \angle{DOC} = {50}^{\circ} \\ \\ \sf \bigstar \: \angle{COB} = {70}^{\circ} \end{array}} \end{gathered}$