Question

law of cosines
which equation correctly uses the law of cosines to solve for the missing side length of △pqr?
triangle with vertices p, q, r: pq = 8, pr = 6, angle at p is 39°, rq = p
law of cosines: ( a^2 = b^2 + c^2 - 2bc cos(a) )
options:
( p^2 = 6^2 + 8^2 - 2(6)(8) cos(39^circ) )
( 6^2 = p^2 + 8^2 - 2(p)(8) cos(39^circ) )
( p^2 = 6^2 + 6^2 - 2(6)(6) cos(39^circ) )
( 8^2 = 6^2 + p^2 - 2(6)(p) cos(39^circ) )

Explanation:

Step1: Recall Law of Cosines

The Law of Cosines is \( a^2 = b^2 + c^2 - 2bc\cos(A) \), where \( a \) is the side opposite angle \( A \), and \( b, c \) are the other two sides.

Step2: Identify sides and angle in \( \triangle PQR \)

• Angle at \( P \) is \( 39^\circ \), so \( A = 39^\circ \).
• Sides adjacent to angle \( P \): \( b = 6 \) ( \( PR \) ), \( c = 8 \) ( \( PQ \) ).
• Side opposite angle \( P \) is \( p \) ( \( QR \) ), so \( a = p \).

Step3: Substitute into Law of Cosines

Substitute \( a = p \), \( b = 6 \), \( c = 8 \), \( A = 39^\circ \) into \( a^2 = b^2 + c^2 - 2bc\cos(A) \):
\( p^2 = 6^2 + 8^2 - 2(6)(8)\cos(39^\circ) \).

Answer:

\( p^2 = 6^2 + 8^2 - 2(6)(8)\cos(39^\circ) \) (the first equation)