In the triangle PQR , PQ = 8, PR = 7 and RQ = 9 Find the size of the largest angle of the triangle. Angle P is largest as it is opposite the largest side. Use the Law of Cosines to find angle P:

`(QR)^2=(PQ)^2+(PR)^2-2(PQ)(PR)cosP`

`81=64+49-2(8)(7)cosP`

`cosP=2/7`

`m/_P=cos^(-1)(2/7)~~73.4^@` .

The area of a traingle can be found if you know two sides and the included angle using `"Area"=1/2 ab sinC`

So `"Area"=1/2(8)(7)sin(73.4)~~26.83` square units.

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Angle P is largest as it is opposite the largest side. Use the Law of Cosines to find angle P:

`(QR)^2=(PQ)^2+(PR)^2-2(PQ)(PR)cosP`

`81=64+49-2(8)(7)cosP`

`cosP=2/7`

`m/_P=cos^(-1)(2/7)~~73.4^@` .

The area of a traingle can be found if you know two sides and the included angle using `"Area"=1/2 ab sinC`

So `"Area"=1/2(8)(7)sin(73.4)~~26.83` square units.

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Largest angle is P with measure 73.4 degrees; the area of the triangle is approximately 26.83 square units

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