75 + 50 +x = 180. ( sum of all angles of in a triangle)
therefore x= 180-(75+50)
x=180-125
x=55°
x and y are supplementary angles therefore
x+y =180
y=180-55. (55 value of x)
y=125
I hope this helps
Answer:
[tex]\qquad \qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: x=55\degree \qquad \: \\ \\& \qquad \:\sf \: y=125\degree \end{aligned}} \qquad \: \\ \\ [/tex]
Step-by-step explanation:
Given that, In triangle PQR
[tex]\qquad\sf \: \angle \: PQR \: = \: 50 \degree \: \\ \\ [/tex]
[tex]\qquad\sf \: \angle \: QPR \: = \: 75 \degree \: \\ \\ [/tex]
[tex]\qquad\sf \: \angle \: PRQ \: = \: x \: \\ \\ [/tex]
and
[tex]\qquad\sf \: \angle \: PRS \: = \: y \: \\ \\ [/tex]
Now, In triangle PQR,
[tex]\qquad\sf \: \angle \: PRS \: is \: an \: exterior \: angle. \: \\ \\ [/tex]
We know, Exterior angle of a triangle is equals to sum of interior opposite angles.
[tex]\qquad\sf \: \angle \: PRS \: = \angle \:PQR \: + \: \angle \:QPR \: \\ \\ [/tex]
[tex]\qquad\sf \: y \: = \: 50\degree \: + \: 75\degree \: \\ \\ [/tex]
[tex]\sf\implies \bf \: y \: = \: 125\degree \: \\ \\ [/tex]
Now, Further, we know that sum of all interior angles of a triangle is supplementary.
So, In triangle PQR
[tex]\qquad\sf \: \angle \:PQR \: + \: \angle \:QRP \: + \: \angle \:QPR \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: 75\degree \: + \: x \: + \: 50\degree \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: 125\degree \: + \: x \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: x \: = \: 180\degree - 125\degree \\ \\ [/tex]
[tex]\sf\implies \sf \: x \: = \: 55\degree \\ \\ [/tex]
Hence,
[tex]\sf\implies \qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: x=55\degree \qquad \: \\ \\& \qquad \:\sf \: y=125\degree \end{aligned}} \qquad \: \\ \\ [/tex]
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Answers & Comments
75 + 50 +x = 180. ( sum of all angles of in a triangle)
therefore x= 180-(75+50)
x=180-125
x=55°
x and y are supplementary angles therefore
x+y =180
y=180-55. (55 value of x)
y=125
I hope this helps
Verified answer
Answer:
[tex]\qquad \qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: x=55\degree \qquad \: \\ \\& \qquad \:\sf \: y=125\degree \end{aligned}} \qquad \: \\ \\ [/tex]
Step-by-step explanation:
Given that, In triangle PQR
[tex]\qquad\sf \: \angle \: PQR \: = \: 50 \degree \: \\ \\ [/tex]
[tex]\qquad\sf \: \angle \: QPR \: = \: 75 \degree \: \\ \\ [/tex]
[tex]\qquad\sf \: \angle \: PRQ \: = \: x \: \\ \\ [/tex]
and
[tex]\qquad\sf \: \angle \: PRS \: = \: y \: \\ \\ [/tex]
Now, In triangle PQR,
[tex]\qquad\sf \: \angle \: PRS \: is \: an \: exterior \: angle. \: \\ \\ [/tex]
We know, Exterior angle of a triangle is equals to sum of interior opposite angles.
[tex]\qquad\sf \: \angle \: PRS \: = \angle \:PQR \: + \: \angle \:QPR \: \\ \\ [/tex]
[tex]\qquad\sf \: y \: = \: 50\degree \: + \: 75\degree \: \\ \\ [/tex]
[tex]\sf\implies \bf \: y \: = \: 125\degree \: \\ \\ [/tex]
Now, Further, we know that sum of all interior angles of a triangle is supplementary.
So, In triangle PQR
[tex]\qquad\sf \: \angle \:PQR \: + \: \angle \:QRP \: + \: \angle \:QPR \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: 75\degree \: + \: x \: + \: 50\degree \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: 125\degree \: + \: x \: = \: 180\degree \\ \\ [/tex]
[tex]\qquad\sf \: x \: = \: 180\degree - 125\degree \\ \\ [/tex]
[tex]\sf\implies \sf \: x \: = \: 55\degree \\ \\ [/tex]
Hence,
[tex]\sf\implies \qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: x=55\degree \qquad \: \\ \\& \qquad \:\sf \: y=125\degree \end{aligned}} \qquad \: \\ \\ [/tex]