The value of x is 67°
<u>Step-by-step explanation:</u>
Given that
PN=LN
NP||MQ
QL bisects <PQM
therefore <PQL=<LQM
NP||MQ and NM is a transversal
<PNL+<LMQ=180°(angles on the same side of the transversal are supplementary)
<PNL+54=180°
<PNL=180-54=126°
Consider ΔPNL
since PN=NL,the triangle is isocelus
<NPL=<NLP=a
<NPL+<NLP+<PNL=180°
a+a+126=180°
2a+126=180
2a=180-126
=54°
a=54/2=27°
consider the point L
<NLP+<PLQ+<MLQ=180°
27+70+<MLQ=180
<MLQ=180-97=83°
consider ΔLQM
<LQM+<LMQ+LMQ=180
<LQM+83+54=180
<LQM=180-(83+54)=180-137=43°
<PQL=43°(since<PQL=<LQM)
considerΔPQL
x+70+<PQL=180°
x+70+43=180°
x+113=180
x=180-113
=67°
The value of x is 67°
Answer:
Probability=2/5
Step-by-step explanation:
Let the line segment be AB of uniform length and x is distance
so
Case 1 min(x,L-x) then
Case 2 min(x,L-x)=L-x then
So Probability
Answer:
D 73º
Step-by-step explanation:
Total angles
a = (sides - 2) * 180
a = (5 - 2) * 180
a = 540
-------------------
Missing angle
x = 540 - (104 + 118 + 96 + 115)
x = 107
a is supplementary to x
m∠A = 180 - 107
m∠A = 73º
