in the circle below, suppose ( moverarc{rqt} = 192^circ ) and ( mangle qrs = 138^circ ). find the following. (a) ( mangle rqt = square^circ ) (b) ( mangle qts = square^circ )

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# Explanation:
## Step1: Find $ m\angle RQT $
A cyclic quadrilateral has opposite angles that sum to $ 180^\circ $. But first, note that the measure of an inscribed angle is half the measure of its intercepted arc. Wait, actually, for cyclic quadrilateral $ QRST $, $ \angle QRS $ and $ \angle QTS $ are related, but for $ \angle RQT $, we can use the fact that the arc $ RQT $ is $ 192^\circ $, so the remaining arc $ RST $ (wait, no, the total circumference is $ 360^\circ $, so the arc $ RT $ (the minor arc) would be $ 360^\circ - 192^\circ = 168^\circ $? Wait, no, maybe better to use cyclic quadrilateral properties. Wait, $ \angle QRS = 138^\circ $, and in cyclic quadrilateral $ QRST $, $ \angle QRS + \angle QTS = 180^\circ $, but for $ \angle RQT $, let's recall that the inscribed angle over arc $ RT $. Wait, the arc $ RQT $ is $ 192^\circ $, so the arc $ RT $ (the major arc? No, minor arc $ RT $ would be $ 360 - 192 = 168^\circ $? Wait, no, maybe I messed up. Wait, the arc $ RQT $ is given as $ 192^\circ $, so the central angle for arc $ RQT $ is $ 192^\circ $, so the inscribed angle over arc $ RT $ (the minor arc) would be half of $ 360 - 192 = 168^\circ $? Wait, no, let's think again. For $ \angle RQT $, it's an inscribed angle intercepting arc $ RT $. The measure of arc $ RT $ is $ 360^\circ - 192^\circ = 168^\circ $? Wait, no, arc $ RQT $ is the arc from $ R $ to $ Q $ to $ T $, so the arc from $ R $ to $ T $ through $ Q $ is $ 192^\circ $, so the minor arc $ RT $ is $ 360 - 192 = 168^\circ $? Wait, no, that can't be. Wait, maybe the arc $ RQT $ is the major arc? No, the problem says $ m\overarc{RQT} = 192^\circ $, so that's the measure of the arc from $ R $ to $ Q $ to $ T $. Then the inscribed angle $ \angle RQT $ intercepts arc $ RT $ (the minor arc). Wait, the measure of an inscribed angle is half the measure of its intercepted arc. So if arc $ RQT $ is $ 192^\circ $, then the minor arc $ RT $ is $ 360 - 192 = 168^\circ $? No, that would make the inscribed angle $ \angle RQT $ half of $ 168^\circ $? Wait, no, maybe I have the arc wrong. Wait, $ \angle RQT $ is an inscribed angle intercepting arc $ RT $. So the measure of $ \angle RQT $ is half the measure of arc $ RT $. The arc $ RT $ (minor arc) is $ 360^\circ - 192^\circ = 168^\circ $? Wait, no, that would mean $ \angle RQT = \frac{1}{2} \times 168^\circ = 84^\circ $? Wait, but let's check the cyclic quadrilateral. $ \angle QRS = 138^\circ $, and in cyclic quadrilateral $ QRST $, $ \angle QRS + \angle QTS = 180^\circ $, so $ \angle QTS = 180 - 138 = 42^\circ $. Wait, maybe part (a) is $ \angle RQT $. Wait, maybe I made a mistake. Wait, the arc $ RQT $ is $ 192^\circ $, so the inscribed angle over arc $ RT $ (minor arc) is $ \frac{1}{2}(360 - 192) = \frac{1}{2}(168) = 84^\circ $. So $ m\angle RQT = 84^\circ $.

## Step2: Find $ m\angle QTS $
In cyclic quadrilateral $ QRST $, opposite angles are supplementary. So $ \angle QRS + \angle QTS = 180^\circ $. Given $ m\angle QRS = 138^\circ $, then $ m\angle QTS = 180^\circ - 138^\circ = 42^\circ $. Alternatively, using the arc: the arc $ QRS $ (wait, no, $ \angle QTS $ intercepts arc $ QRS $. Wait, the arc $ RQT $ is $ 192^\circ $, so the arc $ QRS $ (the arc from $ Q $ to $ R $ to $ S $)? Wait, no, $ \angle QTS $ intercepts arc $ QRS $. Wait, the measure of $ \angle QTS $ is half the measure of arc $ QRS $. Wait, arc $ RQT $ is $ 192^\circ $, so arc $ QTS $ (the other arc) would be $ 360 - 192 = 168^\circ $? No, maybe better to use the cyclic quadrilateral. Since $ QRST $ is cyclic, $ \angle QRS + \angle QTS = 180^\circ $, so $ 138 + \angle QTS = 180 $, so $ \angle QTS = 42^\circ $.

# Answers:
(a) $ m\angle RQT = \boxed{84^\circ} $

(b) $ m\angle QTS = \boxed{42^\circ} $

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QUESTION IMAGE

Question

Explanation:

Step1: Find \( m\angle RQT \)

Step2: Find \( m\angle QTS \)

Answer: