Question

Consider the following figure.A circle, four points on the circle, two interior line segments, and two angles are given.Point R is on the top left.Point S is on the lower left.Point T is on the upper right.Point V is on the right, just below center.Segment R V goes from R down and right to V.Segment S T goes from S up and right to T, intersecting the first segment inside the circle.The angle left of the intersection point, formed below R V and above S T, is labeled 1.The angle under the intersection point, formed below R V and below S T, is labeled 2.Given:m∠1 = 55°mRS$\frown{\hspace{25px}}$ = (5x + 2)°mVT$\frown{\hspace{25px}}$ = x°Find: mRS$\frown{\hspace{25px}}$ in degreesWrite an equation relating m∠1 to mRS$\frown{\hspace{25px}}$ and mTV$\frown{\hspace{25px}}$ in terms of x.55° =°Find x°.x° =°Find mRS$\frown{\hspace{25px}}$ in degrees.mRS$\frown{\hspace{25px}}$ =°

Answer 1

Solution:

Given a circle;

Where

$\begin{gathered} m\angle1=55\degree \\ Arc\text{ RS}=(5x+2)\degree \\ Arc\text{ VT}=x\degree \end{gathered}$

Applying the secant theorem for an interior case,

It states that the measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle.

$m\angle1=\frac{1}{2}(arc\text{ RS}+arc\text{ TV})$

The equation becomes

$55\degree=[\frac{1}{2}((5x+2)+x)]\degree$

Hence,

$55\operatorname{\degree}=\frac{1}{2}[(5x+2)+x]\degree$

Solving for x

$\begin{gathered} 55\degree=\frac{1}{2}(5x+2+x)\degree \\ 55\degree=\frac{1}{2}(6x+2) \\ Crossmultiply \\ 55\degree\times2=6x+2 \\ 110=6x+2 \\ Collect\text{ like terms} \\ 6x=110-2 \\ 6x=108 \\ Divide\text{ both sides by 6} \\ \frac{6x}{6}=\frac{108}{6} \\ x\degree=18\degree \end{gathered}$

Hence, x is 18°

For arc RS

$\begin{gathered} mRS=5x+2=5(18)+2=8=90+2=92\degree \\ mRS=92\degree \end{gathered}$

Hence, mRS is 92°