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:

Step1: Recall circle - triangle property

A triangle inscribed in a circle with one side as the diameter has a right - angle opposite the diameter. In \(\triangle ABC\) with \(AC\) as the diameter of circle \(O\), \(\angle B = 90^{\circ}\) (Thales' theorem).

Step2: Analyze option A

There is no reason for \(AO\) (radius) to be equal to \(AB\) in general.

Step3: Analyze option C

By the triangle inequality theorem, \(AB + BC>AC\) for non - degenerate triangles, and equality \(AB + BC = AC\) holds only when \(A\), \(B\), and \(C\) are collinear, which is not the case here.

Step4: Analyze 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 \(360^{\circ}\).

For the second part:

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 \(QRST\), \(\angle QTS\) and \(\angle QRS\) are opposite angles.

Step2: Calculate \(m\angle QRS\)

Since \(\angle QTS+\angle QRS = 180^{\circ}\) and \(m\angle QTS = 87^{\circ}\), then \(m\angle QRS=180^{\circ}-87^{\circ}=93^{\circ}\)

Answer:

For the first question: B. \(m\angle B = 90^{\circ}\)
For the second question: \(m\angle QRS = 93^{\circ}\)