Question

How to find the values of: a) cos (A-B) b) tan (A+B) while angle is obtuse with sin A = 3/5 and angle B is acute with sin B = 12/13 ?

Answer 1

Below

Explanation:

If angle A is obtuse, then using a right-angled triangle,

`sin A=3/5`
`cos A=-4/5` (cos is negative in the second quadrant)
`tan A=-3/4` (tan is negative in the second quadrant)

If angle B is acute, then using a right-angled triangle,

`sin B=12/13`
`cos B=5/13`
`tan B=12/5`

a)
`cos(A-B)=cosAcosB+sinAsinB`
`=(-4/5)(5/13)+(3/5)(12/13)`
`=-4/13+36/65`
`=16/65`

b)
`tan(A+B)=(tanA+tanB)/(1-tanAtanB)`
`=(-3/4+12/5)/(1-(-3/4)(12/5))`
`=(33/20)/(1+9/5)`
`=33/20times5/14`
`=33/56`