(I don't know how to upload pictures but if you search these questions up it will be the same diagram) | | | 1.Lines XQ and XR are tangent to a circle, as shown below. If angle QTA = 47 and angle RUA = 65, then find angle QXR, in degrees. | | 2.The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ | | 3.Quadrilateral $ABCD$ is an isosceles trapezoid, with bases $\overline{AB}$ and $\overline{CD}.$ A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base $\overline{AB}$ is $2x,$ and the length of base $\overline{CD}$ is $2y.$ Prove that the radius of the inscribed circle is $\sqrt{xy}.$
1. The answer is 78 degrees.
The two tangents from a point to a circle are congruent, and they form an angle of 180 degrees with the line that passes through the center of the circle and that point.
In the diagram, we know that angle QTA is 47 degrees, and angle RUA is 65 degrees. These two angles are supplementary angles, so they add up to 180 degrees. Therefore, angle QXR is 180 - (47 + 65) = 78 degrees.
2.
Let $O$ be the circumcenter of $\triangle WXY,$ and let $A$ be the midpoint of $\overline{OX}.$ Then $WA=WX/2=3$ and $OA=OX/2=7/2,$ so by the Pythagorean Theorem, $OW^2=OA^2+WA^2=37/4,$ and $OW=\sqrt{37}/2.$
Let $r$ be the radius of $\odot(WXY).$ Then $XZ=XY-YZ=14-YZ.$ Note that $WZ$ is parallel to the tangent at $X,$ so $\angle WZX=\angle X,$ and $\angle WXA=90^\circ.$ Thus $\triangle WZX\sim\triangle XAO,$ so $WZ/XA=ZX/OA,$ or $WZ=ZX\cdot OA/XA.$ Since $\triangle WXY$ is right-angled at $W,$ we have $WX^2+XY^2=WY^2,$ so $WY=4\sqrt{10},$ and $r=WY/2=2\sqrt{10}.$ Hence $OA^2=r^2+AX^2=40+3^2=49,$ so $OA=7.$ Therefore $WZ=(14-YZ)\cdot 7/3$ and $OZ=OW+WZ=\sqrt{37}/2+7(14-YZ)/3.$
Since $OZ$ is a radius of $\odot(WXY),$ we have $OZ=r,$ which gives 37/2+7(14−��)/3=210.37/2+7(14−YZ)/3=210. Solving for $YZ$ gives $YZ = \frac{63}{5}$.
