Question

1. triangle abc is inscribed in circle o as shown. $overline{ac}$ is a diameter of circle o. which relationship must be true? a $ao = ab$ b $mangle b = 90^{circ}$ c $ab + bc = ac$ d $mangle a+mangle b+mangle c = 360^{circ}$ 4. quadrilateral qrst is inscribed in circle f, as shown. the measure of $angle qts$ is $87^{circ}$. what is $mangle qrs$?

Explanation:

Question 3

Step1: Recall circle - inscribed triangle property

An angle inscribed in a semi - circle is a right angle. Since $\overline{AC}$ is a diameter of circle $O$ and $\angle B$ is an inscribed angle that intercepts the semi - circle, by the inscribed - angle theorem for semi - circles.

Step2: Analyze each option

• Option A: $AO$ is a radius and $AB$ is a chord. There is no reason for $AO = AB$ in general.
• Option B: Since $\angle B$ is inscribed in a semi - circle (the arc $AC$ is a semi - circle), $m\angle B=90^{\circ}$.
• Option C: By the triangle inequality, $AB + BC>AC$ for non - degenerate $\triangle ABC$.
• Option D: The sum of the interior angles of a triangle is $180^{\circ}$, so $m\angle A + m\angle B+m\angle C = 180^{\circ}$, not $360^{\circ}$.

Step1: Recall the property of a cyclic quadrilateral

The opposite angles of a cyclic quadrilateral (a quadrilateral inscribed in a circle) are supplementary. In cyclic quadrilateral $QRST$ inscribed in circle $F$, $\angle QTS$ and $\angle QRS$ are opposite angles.

Step2: Use the supplementary - angle relationship

If two angles are supplementary, the sum of their measures is $180^{\circ}$. Given $m\angle QTS = 87^{\circ}$, let $m\angle QRS=x$. Then $x + 87^{\circ}=180^{\circ}$.

Step3: Solve for $x$

$x=180^{\circ}-87^{\circ}=93^{\circ}$.

Answer:

B. $m\angle B = 90^{\circ}$

Question 4