19.2 I think. Though I'm not sure because it isn't a whole number
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:
the third one, y = 1/2x + 3
You could use Pythagoras Theorem to solve this.
All the points are equidistant from the centre of line AB, therefore it is just the same triangle but rotated. I hope this help but I can show you on paper if you don’t understand that answer :)