How do you solve #13^2=8^2+6^2-2(8)(6)cosA#?

2 Answers
Sep 23, 2017

Answer:

#angleA=136^@#

Explanation:

#"isolate "cosA" by evaluating the numeric values"#

#rArr169=64+36-96cosA#

#rArr-96cosA=69#

#rArrcosA=-69/96=-23/32#

#rArrA=cos^-1(-23/32)=136^@#

Sep 23, 2017

Answer:

#rArrA = 136^o +2kpi" "# where #k in NN#
#" "#
Or
#rArr A=224^o +2kpi" "#where #k in NN#

Explanation:

#13^2-8^2-6^2=-2 (8)(6)cosA#
#" "#
#rArr169-64-36 =-2 (8)(6)cosA#
#" "#
#rArr169-64-36 = -96cosA#
#" "#
#rArr 69 = -96cosA#
#" "#
#rArrcosA = -69/96#
#" "#
#rArr CosA = -0.71875#
#" "#
Since #cosA <0" "#then #A^o in # quadrants 2&3
#" "#
#rArrA = 136^o +2kpi" "# where #k in NN#
#" "#
Or
#rArr A=224^o +2kpi" "#where #k in NN#