Guest Apr 17, 2019

edited by
Guest
Apr 17, 2019

edited by Guest Apr 17, 2019

edited by Guest Apr 17, 2019

edited by Guest Apr 17, 2019

edited by Guest Apr 17, 2019

#1**+1 **

Thank you for trying before you post on this forum. I appreciate it. (I am really tired of all the people just being lazy and not trying at all and posting on here).

As for your question.

To find angle \(Q\) and angle \(P\), you need to first solve for \(y\) . We first solve for the unknown angle \(\angle QRP\). We see that this angle and the angle \(\angle QRS\) add up to 180º (Straight Line). As you already did, 180º-150º is 27º.

We know that the sum of the angles in a triangle add up to 180º, so we can now make a linear equation to solve for \(y\). The Equation is as follows:

\((3y+5)+(2y-7)+27=180\)

We then solve for y:

\((3y+5)+(2y-7)+27=180 \)

\( 3y+5+2y-7+27=180\) -- Remove Parentheses

\(5y+25=180\) -- Combine like terms

\(5y=155\) -- Isolate the Variable

\(y=31\)

We can now solve for each of the angles.

\(m\angle Q= 3(31)+5\\m\angle Q= 93+5\)

\(m\angle Q= \) \(\boxed{98}\)

\(m\angle P=2(31)-7\\m\angle P=62-7\\m\angle P=\boxed{55}\)

\(m\angle QRP=\boxed{27}\) (We already solved this.)

**This was my first time using LaTeX, please forgive me. :)**

Bxtterman Apr 17, 2019