Question

Can you help me find the angles of x°, R, Q, and P?

Answer 1

SOLUTION

Using the theorem

The sum of angles in a triangle is 180°

Then we have

$P^0+Q^0+R^0=180^0$

from the diagram in the questions

$\begin{gathered} P=x-11 \\ Q=3x+6 \\ R=x \end{gathered}$

Substituting the parameters we have

$\begin{gathered} (x-11)^0+(3x+6)^0+x^0=180^0 \\ \text{ Remove the parenthesis} \\ x-11+3x+6+x=180^0 \\ \text{ Rearrange the expression} \\ x+3x+x+6-11=180^0 \end{gathered}$

Simplify the equation above

$\begin{gathered} 5x-5=180^0 \\ 5x=180+5 \\ 5x=185 \\ \text{Divide both sides by 5} \\ x=\frac{185}{5} \end{gathered}$

Therefore

$x=39^0^{}$

Hence

The value of x is 39°