I'm pretty sure it's figure A
That’s 4- 1 and 3/4. The answer is 2 and 1/4
Answer:
x=27°
Step-by-step explanation:
we know that
<u><em>The Triangle Exterior Angle Theorem</em></u>, states that: An exterior angle of a triangle is equal to the sum of the opposite interior angles
step 1
Find the measure of angle PLM
we know that
m∠PLM+m∠LPM=m∠PMN ----> by Triangle Exterior Angle Theorem
we have
m∠LPM=x°
m∠PMN=2x+72°
substitute
m∠PLM+x=2x+72°
m∠PLM=2x-x+72°
m∠PLM=x+72°
step 2
Find the measure of angle x
we know that
3x+m∠PLM=180° ----> by supplementary angles (form a linear pair)
we have
m∠PLM=x+72° (see step 1)
substitute
3x+x+72°=180°
4x=180°-72°
4x=108°
x=27°
Angle QPS is an inscribed angle. Arc QS is the intercepted arc of that angle. The rule is that the intercepted arc is twice its angle measure. P measures 20 degrees, so arc QS measures 40 degrees. It just so happens that angle QTS is ALSO an inscribed angle intercepting arc QS. So angle QTS measures the same as angle P, 20 degrees. That's a. For b., we already stated the rule and figured out that minor arc QS is 40 degrees. For c., major arc QTS is 360 (the measure around the outside of a circle...EVERY circle in the world) minus the minor arc of 40. So major arc QTS measures 360 - 40 which is 320 degrees. There you go!
