Answer:
m∠QPR = 80°
Explanation:
Look at the picture.
We know: the sum of the measures of the angles of the triangle is equal to 180 °.
Therefore
In ΔTQR we have the equation:
subtract 130° from both sides
![\alpha+\beta=50^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b7jezticlwsf736zsq9g9rdel3tlo6yqos.png)
In ΔPQR we have the equation:
![2\alpha+2\beta+m\angle QPR=180^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zbkbaqavdf8wogqw1gapbhkvyu7mdpmyzc.png)
![2(\alpha+\beta)+m\angle QPR=180^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r4iuz6sdfam7ijghbr49jr3x4qxnviy8rq.png)
Substitute from the first equation
![\alpha+\beta=50^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b7jezticlwsf736zsq9g9rdel3tlo6yqos.png)
![2(50^o)+m\angle QPR=180^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1zr6ozeox0aw4zs5tb7krqw78c6m3916l8.png)
subtract 100° from both sides
![m\angle QPR=80^o](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fg2ewhkcp2g2k21og4aeygb0wwu0qvs0rs.png)