Asked by heather
In triangle PQR, m<P=53 degrees, PQ=7.4, and PR=9.6. What is m<R to the nearest degree
Answers
Answered by
Damon
first what is QR which is p?
law of cosines
p^2 = q^2 + r^2 - 2 p q cos P
p^2 = 9.6^2 + 7.4^2 - 2(7.4)(9.6)cos 53
solve for p
then law of sines
sin R/7.4 = sin 53/ p
law of cosines
p^2 = q^2 + r^2 - 2 p q cos P
p^2 = 9.6^2 + 7.4^2 - 2(7.4)(9.6)cos 53
solve for p
then law of sines
sin R/7.4 = sin 53/ p