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In the triangle PQR , PQ = 8, PR = 7 and RQ = 9 Find the size of the largest angle of...

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saj-94 | (Level 1) Salutatorian

Posted July 24, 2013 at 6:16 PM via web

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In the triangle PQR , PQ = 8, PR = 7 and RQ = 9

Find the size of the largest angle of the triangle. 

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llltkl | College Teacher | (Level 3) Valedictorian

Posted July 24, 2013 at 7:08 PM (Answer #1)

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The law of sines require that the angle opposite to the largest side must be the largest angle.

Here, largest side is QR.

So, `angleP` must be the largest.

It can be found out using the law of cosines. Thus,

`cosangleP=(8^2+7^2-9^2)/(2*8*7)=32/112=0.2857`

`rArr angleP=arccos(0.2857)=73.4^o`

Area of the triangle can be obtained by Heron's formula.

Here `s=(8+7+9)/2=12 ` units

Area A=`sqrt(12*(12-8)(12-7)(12-9))` sq. units

`=sqrt(12*4*5*3)=sqrt720 ` sq. unit

`=26.83 ` sq. units

Therefore, the measure of the largest angle of the triangle is 73.4 degrees and its area is 26.83 sq. units.

Sources:

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 25, 2013 at 12:49 PM (Answer #2)

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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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