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's profile pic

Posted on (Answer #1)

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,


`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.

embizze's profile pic

Posted on (Answer #2)

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




`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.


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


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