Question

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.​

Answer 1

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...

Answer 2

$\LARGE\sf {\underline{\underline\green{Required \:  Answer:-}}}$

In ∆ABC, exterior ∠CBX at B and adjacent interior ∠CBA from a linear pair.

$\sf{∠CBX + ∠CBA = 180⁰}$

$\sf\implies{70⁰ + y⁰=180⁰}$

$\sf\implies{y⁰=180⁰-70⁰=110⁰}$

$\boxed{ \therefore{ \sf{y⁰=110⁰}}}$

Again, ∠BCY is exterior of ∆ABC at C, and ∠CAB and ∠CBA are interior opposite angles.

$\sf{\therefore{x⁰=10⁰+y⁰}}$

$\sf\implies{x⁰=10⁰+110⁰.(∵y⁰=110⁰)}$

$\boxed{ \therefore{x⁰= 120⁰}t}$

Hence, we have x⁰ = 120⁰ and y⁰ = 110⁰.

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