Question

Find the value of x in the attached fig.​

Answer 1

The value of x for the given diagram or figure is 32°.

Explanation:

Here, we are given with a diagram that has a straight line. This straight line is divided into two different parts. There are two 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 the 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.

Solution:

Angle POR and QOR form a linear pair

Angle POR + Angle QOR = 180 [Linear pair axiom]

Substitute all the angles values in equation then

⇛(2x+60)° + (3x-40)° = 180°

Open all brackets in LHS.

⇛2x + 60° + 3x - 40° = 180°

Add all the variables and constants separately.

⇛5x + 20° = 180°

Shift the number 20° from LHS to RHS, changing it's sign.

⇛5x = 180° - 20°

Subtract the values on RHS.

⇛5x = 160°

Shift the number 5 from LHS to RHS.

⇛x = 160°/5°

Simplify the fraction to get the value of x.

⇛x = 32°

Therefore, x = 32°

Answer: Hence, the value of x is 32°

EXPLORE MORE:

Now, let's find the value of each angles separately. By putting the value of x in their places.

Angle POR

⇛(2x + 60)° = (2*32 + 60)° = (64 + 60)° = (124)°

So, Angle POR = 134°

Next,

Angle QOR

⇛(3x-40)° = (3*32 + 40)° = (96 - 40)° = (56)°

Angle QOR = 56°

Please let me know if you have any other questions.

Answer 2

`\boxed{\boxed{\sf x = 32{}^( \circ)}}`

Explanation:

The given pair of angles is Linear pair.

[*If the adjacent angles are drawn in such a way that they form straight line , then the angles are known as linear pair*]

We know that their sum is equal to 180°.

So,

`\rm\angle{POR} + \angle{QOR} = \tt180 {}^( \circ)`

Note that,

[POR = (2x+60)°]

AND,

[QOR = (3x-40)°]

`\tt \implies(2x + 60) {}^( \circ) + (3x - 40){}^( \circ) = 180{}^( \circ)`

`\tt \implies2x + 60 + 3x - 40 = 180{}^( \circ)`

Now find the value of x.

Steps:

`\tt \implies2x + 60 + 3x - 40 = 180{}^( \circ)`

The LHS may be rewritten as,

`\tt \implies2x + 3x + 60 + ( - 40) = 180{}^( \circ)`

So, Combine like terms :

`\tt \implies(2x + 3x) + 60 - 40 = 180{}^( \circ)`

`\tt \implies3x + (60 - 40) = 180 {}^ \circ`

• Subtract the bracketed portion:

`\tt \implies5x + 20 = 180{}^( \circ)`

Now, Subtract 20 from both sides:

`\tt \implies5x + 20 - 20 = 180{}^( \circ) - 20`

• Simplify the RHS and LHS:

`\tt \implies5x + 0 = 160{}^( \circ)`

`\tt \implies5x = 160{}^( \circ)`

Divide both sides by 5:

`\tt \implies \cfrac{5x}{5} = \cfrac{160{}^( \circ) }{5}`

• Cancel the RHS and LHS:

`\tt \implies \cfrac {\cancel5 {}^(1) x}{ \cancel5 {}^(1) } = \cfrac{ \cancel{160} {}^( \circ) \: {}^{32{}^( \circ) } }{ \cancel5 {}^(1) }`

`\tt \implies1x = 32{}^( \circ)`

`\tt \implies{x} = 32{}^( \circ)`

Hence, the value of x would be 32°.

`\rule{225pt}{2pt}`

I hope this helps!

Let me know if you have any questions.