Q8. - Geometry

Question

$\overline{TR}$ and $\overline{TS}$ are tangents to $\odot$$O$ from $T;$

                                 $m\angle RTS=36$

$\left(a\right).$ Copy the diagram. Draw $\overline{RS}$ and find $m\angle TSR$ and $m\angle TRS.$

$\left(b\right).$ Draw radii $\overline{OS}$ and $\overline{OR}$ and find $m\angle ORS$ and $m\angle OSR$.

$\left(c\right).$ Find $m\angle ROS$.

$\left(d\right).$ Does your result in part $\left(c\right)$ support one of your conclusions about angles in Classroom Exercise $5?$ Explain.

Step-by-Step Solution

Answer

a. The value of, $m\angle TSR=72°$ and $m\angle TRS=72°$.

b. The value of, $m\angle ORS=m\angle OSR=18°$.

c. The value of, $m\angle ROS=144°$.

Part a. Step 1. Given information.

Given:

$TR$ and $TS$ are tangents to the circle at $O$ from $T$.

$m\angle RTS=36°$.

Step 2. Tangents from a same point towards the circle are equal.

$TR=TS$

In $\Delta RTS$,

$m\angle RTS=36°$. 

Since $TR=TS$,

$\angle TSR=\angle TRS=x$ (angles opposite to equal sides are equal).

Again,

Step 3. Apply angle sum property of a triangle).

since $m\angle RTS+m\angle TRS+m\angle TSR={180}^0$

$180{}^{\circ }=36{}^{\circ }+x+x$,

$180{}^{\circ }=36{}^{\circ }+2x$,

$x=72°$

$\therefore m\angle TSR=m\angle TRS=72°$.

Therefore, the measures are: $m\angle TSR=72°$ and $m\angle TRS=72°$.

Part b. Step 1. Given information.

Given:

TR and $TS$ are tangents to the circle at $O$ from $T$.

$m\angle RTS=36°$.

Step 2. Apply angle sum property of a quadrilateral.

In quadrilateral $RTSO$,

$\angle RTS+\angle TSO+\angle SOR+\angle ORT=360°$

$\angle TRO=\angle TSO=90°$ (a tangent form an angle of $90°$ with the surface of the circle.)

Step 3. By plugging the values.

we have,

$36{}^{\circ }+90{}^{\circ }+90{}^{\circ }+\angle ROS=360°$

$216{}^{\circ }+\angle ROS=360°$

$m\angle ROS=144°$ 

Now in $\Delta SOR$,

$RO=SO$ (radii of the circle),

$\angle ORS=\angle OSR=x$ (angle opposite to equal sides are equal).

Step 4. Now, apply angle sum property of a triangle.

$\angle ROS+\angle ORS+\angle OSR=180°$

$144{}^{\circ }+x+x=180°$,

$2x=36°$

$x=18°$

Therefore, value of, $m\angle ORS=m\angle OSR=18°$.

Part c. Step 1. Given information.

Given:

$TR$ and $TS$ are tangents to the circle at $O$ from T.

$m\angle RTS=36°$.

Step 2. Apply angle sum property of a quadrilateral.

In quadrilateral $RTSO$,

$\angle RTS+\angle TSO+\angle SOR+\angle ORT=360°$

$\angle TRO=\angle TSO=90°$ (a tangent form an angle of $90°$ with the surface of the circle.)

Step 3. By plugging the values.

we have,

$36{}^{\circ }+90{}^{\circ }+90{}^{\circ }+\angle ROS=360°$,

$216{}^{\circ }+\angle ROS=360°$,

$m\angle ROS=144°$.

Therefore, the value of $m\angle ROS=144{}^{\circ }.$