Part 1) In triangle PQR, mP= 43, PQ=7.5 and PR= 8.4. What is mR to the nearest degree?
we know that
the law of cosines establishes that
c²=a²+b²-2*a*b*cos C
in this problem
c=QR
a=PQ=7.5
b=PR=8.4
C=angle P=43°
c²=a²+b²-2*a*b*cos C----> c²=7.5²+8.4²-2*7.5*8.4*cos 43°
c²=34.66--------> c=5.89
QR=5.9
applying the law of sines
7.5/sin R=5.9/sin 43-----> 7.5*sin 43=5.9*sin R---> sin R=7.5*sin 43/5.9
sin R=0.8669---------> R=arc sin (0.8669)----> R=60.11°----> R=60°
the answer Part 1) is the option
D. 60
Part 2) In triangle ABC mA=43, mB=62 and BC=22 in. What is AB to the nearest tenth of an inch?
we know that
∠C=180-(62+43)-----> ∠C=75°
Applying the law of sines
22/sin 43=AB/sin 75------> AB=22*sin 75/ sin 43-----> AB=31.16 in
AB=31.2 in
the answer Part 2) is
D.31.2
