Question

Triangle P R Q is shown. Angle R P Q is 39 degrees. The length of P R is 6, the length of R Q is p, and the length of P Q is 8.
Law of cosines: a2 = b2 + c2 – 2bccos(A)

Which equation correctly uses the law of cosines to solve for the missing side length of TrianglePQR?

62 = p2 + 82 – 2(p)(8)cos(39°)
p2 = 62 + 82 – 2(6)(8)cos(39°)
82 = 62 + p2 – 2(6)(p)cos(39°)
p2 = 62 + 62 – 2(6)(6)cos(39°)

Answer 1

According to law of cosines the length of RQ can be written as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$.

Given the length PR is 6 , the length of RQ is p, the length of PQ is 8 and the angle RPQ is 39 degrees.

A length of the triangle can be written as according to law of cosines if sides are given and one angle is $a^{2} =b^{2} +c^{2} -2bccos(A)$

We have to just put the values in the above equation.

as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$.

p is the side opposite to angle given , the length of other sides are 6 and 8 and angle is 39 degrees.

Hence the side can be written as according to law of cosines if the angle is 39 degrees is  as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$.

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