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