I need help with this(I know the law of sines and law of cosines). Not sure how to apply that here.

Question

The lengths of segments PQ and PR are 8 inches and 5 inches, respectively, and they make a 60-degree angle at P. Find the third side of another triangle that has a 5-inch side, an 8-inch side, and the same area as triangle PQR.

Answer 1

a=PQ=8 in., b=QS=5 in., P=angle C=60 degrees

c^2= a^2+b^2 -2abCosC

c^2=64+25-2(8)(5)Cos60=89-80(1/2)=49

c=QS=sqr49= 7 inches

area = hb/2=5h/2=2.5h where h=height can be found from 2 adjoining right triangles= 34.62/5= 6.924

area= 2.5(6.924= 17.32 ln^2

or most easily calculated

area = 1/2 x b x c xsinC= .5(8)(5)sin60= 20sqr3/2 = 10sqrt3= about 17.32 in^2

or use Heron's formula

area =sqr(s(s-a)(s-b)(s-c)) where s=semi- perimeter = (8+7+5)/2 = 10

= sqr(10x2x3x5) = sqr300= 10sqr3 in^2 = about 17.32 in^2

sinB/5=sinC/c=sqr3/2/7

sinB=5sqr3/14

angle B= sin^-1(5sqr3/14)= about 38.21 degrees

angle A=180-60-38.21= about 81.79 degrees

all sides, angles and areas are the same for both triangles

but you're looking for triangle PQR compared to QPR

still, if the Area is the same, the 3rd side is still 7 inches

and all sides and angles are the same, just rotate the triangle

to switch the angles and sides