Question

1. in △rst below, u is a point on rt such that su is an angle - bisector of ∠rst. what is m∠r? f. 43.75° g. 50° h. 70° j. 90° k. 100°

Explanation:

Step1: Apply angle - bisector property

Since $\overrightarrow{SU}$ is the angle - bisector of $\angle RST$, $\angle RSU=\angle TSU = x^{\circ}$.

Step2: Use triangle - angle sum theorem

In $\triangle RSU$, the sum of interior angles is $180^{\circ}$. So, $\angle R+ \angle RSU+\angle RUS = 180^{\circ}$. In $\triangle TSU$, $\angle T+\angle TSU+\angle TUS = 180^{\circ}$, and $\angle RUS+\angle TUS = 180^{\circ}$. In $\triangle RST$, $\angle R+\angle RST+\angle T=180^{\circ}$, where $\angle RST = 2x^{\circ}$, $\angle T=(x + 5)^{\circ}$, and in $\triangle RSU$, $\angle RUS = 180-(3x)^{\circ}$.
In $\triangle RST$, we have $\angle R+2x+(x + 5)=180$. Also, in $\triangle RSU$, $\angle R+x+(180 - 3x)=180$. Simplifying $\angle R+x+180 - 3x=180$ gives $\angle R-2x = 0$, so $\angle R = 2x$.
Substitute $\angle R = 2x$ into $\angle R+2x+(x + 5)=180$. Then $2x+2x+x + 5=180$.
Combining like - terms, $5x+5 = 180$.
Subtract 5 from both sides: $5x=175$.
Divide both sides by 5: $x = 35$.
Since $\angle R = 2x$, then $\angle R=70^{\circ}$.

Answer:

H. $70^{\circ}$