let sinA=5/13 ,cosA>0 and tanB=4/3 cosB=0 find the exact value of cos(A-B0

Question

i need help

Answer 1

sine(a)=5/13=0.384615384

a=invers sine90.384615384)=22.61986495 deg

tan(b)=4/3=1.33333333333

b=invers tan(1.33333333)=53.13010235

a-b=-30.5102374 deg

kosine(-30.5102374)=0.861538461

Answer 2

let sinA=5/13 ,cosA>0 and tanB=4/3 cosB=0 find the exact value of cos(A-B).

This is wrong where it says cosB = 0, so that part is being ignored (not needed anyway).

sinA = 5/13.

Therefore cosA = 12/13  (use a right-angled triangle and pythagoras' theorem)

tanB = 4/3,

Therefore, cosB = 3/5, sinB = 4/5   (again, use right-angled triangle)

cos(A + B) = cosA.cosB + sinA.sinB = (12/13)(3/5) + (5/13)(4/5) = 36/65 + 4/13 = 56/65

Answer: cos(A - B) = 56/65