Answer:
1. Given
m<OPQ = 140°
m<POM = 105°
a. let be <OPQ and <OPM be supplementary angles, thus
m<OPQ + m<OPM = 180°
140° + m<OPM = 180°
m<OPM = 180° - 140°
m<OPM = 40°
use definition of a triangle
we have,
m<POM + m<OPM + m<M = 180°
105° + 40° + m<M = 180°
145° + m<M = 180°
m<M = 180° - 145°
b. let be <POM and <NOP be supplementary angle, thus,
m<POM + m<NOP = 180°
105° + m<NOP = 180°
m<NOP = 180° - 105°
c. given
<POM = 105°
<OPM = 40°
<M = 35°
<POM has the largest angle measure, thus by applying greater angle/greater side theorem
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Answers & Comments
Answer:
1. Given
m<OPQ = 140°
m<POM = 105°
a. let be <OPQ and <OPM be supplementary angles, thus
m<OPQ + m<OPM = 180°
140° + m<OPM = 180°
m<OPM = 180° - 140°
m<OPM = 40°
use definition of a triangle
we have,
m<POM + m<OPM + m<M = 180°
105° + 40° + m<M = 180°
145° + m<M = 180°
m<M = 180° - 145°
m<M = 35°
b. let be <POM and <NOP be supplementary angle, thus,
m<POM + m<NOP = 180°
105° + m<NOP = 180°
m<NOP = 180° - 105°
m<NOP = 75°
c. given
<POM = 105°
<OPM = 40°
<M = 35°
<POM has the largest angle measure, thus by applying greater angle/greater side theorem
The longest side of ∆OPM is PM
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