Question #2ec24

Question

1082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T15:05:26

Given
#sinA=12/13 and A" is not in 1st quadrant"#

So #A# must be in 2nd quadrant

Hence #A=sin^-1(12/13)=112.62^@#

Again

#cosB=3/5 and B" is not in 1st quadrant"#

So #B# must be in 4th quadrant

Hence #A=cos^-1(3/5)=306.87^@#

So #A+B=112.62+306.87=419.49^@#

Now #(419.49-360)^@=59.49^@#

Hence #A+B-># In the first quadrant.