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 ?

1 Answer
Jul 27, 2018

Below

Explanation:

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

sin A=3/5sinA=35
cos A=-4/5cosA=45 (cos is negative in the second quadrant)
tan A=-3/4tanA=34 (tan is negative in the second quadrant)

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

sin B=12/13sinB=1213
cos B=5/13cosB=513
tan B=12/5tanB=125

a)
cos(A-B)=cosAcosB+sinAsinBcos(AB)=cosAcosB+sinAsinB
=(-4/5)(5/13)+(3/5)(12/13)=(45)(513)+(35)(1213)
=-4/13+36/65=413+3665
=16/65=1665

b)
tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A+B)=tanA+tanB1tanAtanB
=(-3/4+12/5)/(1-(-3/4)(12/5))=34+1251(34)(125)
=(33/20)/(1+9/5)=33201+95
=33/20times5/14=3320×514
=33/56=3356