Question

HELP ME PLEASE FIND THE ANSWER​

Answer 1

$\begin{gathered}\frak{Given}\begin{cases}\sf{\:\ \: \angle \: ACB = 60^{0}}\\{ \: \: \: \sf \angle \: ABP =100^{0}}\end{cases}\end{gathered}$

To find

• x and y

Step by step explanation

$\sf{\angle ABP+ \angle y\ =180° \ \ \red{ by \ linear \ pair}}$

$\sf{\Longrightarrow 100°+y= 180⁰ }$

$\sf{\Longrightarrow y= 180-100 }$

$\sf{\Longrightarrow y= 80⁰ }$

In ∆ABC , We have

• $\sf{ACB=80° \ and\ y=60° }$

$\sf{\Longrightarrow ACB + x+ y= 180° \: \: \: \: \: \: \red{Angle\ sum \ property }}$

$\sf{\Longrightarrow 60 + x+ 80= 180° }$

$\sf{ \Longrightarrow x+140= 180° }$

$\sf{\Longrightarrow x=180-140 }$

$\sf{\Longrightarrow x=40° }$

• x= 40⁰
• y=80⁰

Answer 2

$\sf \pmb{Answer :}$

$\begin{gathered}\frak{Given}\begin{cases}\sf{\:\ \: \angle \: ACB = 60^{0}}\\{ \: \: \: \sf \angle \: ABP =100^{0}}\end{cases}\end{gathered}$

To find

x and y

Step by step explanation

$\sf{\angle ABP+ \angle y\ =180° \ \ \red{ by \ linear \ pair}}$

$\sf{\Longrightarrow 100°+y= 180⁰ }$

$\sf{\Longrightarrow y= 180-100 }$

$\sf{\Longrightarrow y= 80⁰ }$

In ∆ABC , We have

$\sf{ACB=80° \ and\ y=60° }$

$\sf{\Longrightarrow ACB + x+ y= 180° \: \: \: \: \: \: \red{Angle\ sum \ property }}$

$\sf{\Longrightarrow 60 + x+ 80= 180° }$

$\sf{ \Longrightarrow x+140= 180° }$

$\sf{\Longrightarrow x=180-140 }$

$\sf{\Longrightarrow x=40° }$

x= 40⁰

y=80⁰