Please help!!! I should complete my excercise!!!
The measurements of some of the angles are indicated. Find the values of x⁰ and y⁰.
please answer properly with Explanation, it's very very important.
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Please help!!! I should complete my excercise!!!
The measurements of some of the angles are indicated. Find the values of x⁰ and y⁰.
please answer properly with Explanation, it's very very important.
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HeRe Is UR AnS:
70 + y = 180° ( angle on straight line)
or, y= 110°
10° + y + k (let) = 180° (sum of angles of interior traingle)
or,
10 + 110+ k = 180
or, k = 60
Now,
k+x= 180° (angles on straight line)
or,60+ x =180°
or,x=120°
x=120° and y=110°
STudY WeLL...
Verified answer
[tex]\LARGE\sf {\underline{\underline\green{Required \: Answer:-}}}[/tex]
In ∆ABC, exterior ∠CBX at B and adjacent interior ∠CBA from a linear pair.
[tex] \sf{∠CBX + ∠CBA = 180⁰}[/tex]
[tex] \sf\implies{70⁰ + y⁰=180⁰}[/tex]
[tex] \sf\implies{y⁰=180⁰-70⁰=110⁰}[/tex]
[tex] \boxed{ \therefore{ \sf{y⁰=110⁰}}}[/tex]
Again, ∠BCY is exterior of ∆ABC at C, and ∠CAB and ∠CBA are interior opposite angles.
[tex]\sf{\therefore{x⁰=10⁰+y⁰}}[/tex]
[tex]\sf\implies{x⁰=10⁰+110⁰.(∵y⁰=110⁰)}[/tex]
[tex] \boxed{ \therefore{x⁰= 120⁰}t}[/tex]
Hence, we have x⁰ = 120⁰ and y⁰ = 110⁰.
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