1.

In ∆PQR, PQ = 10 cm, QR = 12cm, PR = 8 cm, find the biggest and the smallest angle of the triangle.

Answer»

In ∆ PQR 12 cm > 10 cm > 8 cm 

∴ QR > PQ > PR 

∴ ∠ P > ∠ R > ∠ Q

The biggest angle is ∠P and the smallest angle is ∠Q.

Biggest angle is P which is opposite to longest side QR.

cos P = [( PQ^2+PR^2-QR^2)/(2*PQ*PR)]^0.5

=> P = cos ^-1(0.125)=82,8 degree

Smallest angle is Q which is opposite to shortest side PR.

cos Q = [( PQ^2+QR^2-PR^2)/(2*PQ*QR)]^0.5

=> Q = cos ^-1(0.75)=41,4degree


Discussion

No Comment Found

Related InterviewSolutions