Question

Determine if it is possible to draw ∆PQR with side PQ=11cm,QR=5cm,PR=5cm​

Answer 1

Answer:

In a triangle PQR, PQ =8cm and QR= 7cm. The area of the triangle is 17cm^ 2. Calculate the two possible values of <PQR. One I found 37.3 deg, how to find the 2nd one?

Area of PQR = (PQ*QR/2)*sin PQR, or

17 = (8*7/2)*sin PQR, or

sin PQR = 17*2/(8*7) = 0.607142857, or

<PQR = 37.38319842 deg.

We can find PR^2 = PQ^2+QR^2–2*PQ*QR cos PQR

= 8^2+7^2–2*8*7*cos 37.38319842

= 64+49–112*0.794592695

= 113–88.99438184

= 24.00561816, or

PR = 4.899552853 cm

Area of PQR = 17 = (QR*QR/2)*sin PRQ

=(7*4.899552853/2)*sin PRQ, or

sin PRQ = 17/(7*4.899552853/2) = 0.991344108

<PRQ = arc sin 0.991344108 = 82.45590259 deg

<QPR = 180 - 82.45590259 - 37.38319842 = 60.16089899 deg

<QPR = 60.16089899 deg and <PRQ = 82.45590259 deg

Another value of <PQR = 180 -37.38319842 = 142.6168016 deg.