For Positive Acute Angles A and B it is known that sin A = 21/29 and tan B = 28/45. Find the value of cos (A-B) in simplist form.

Question

For Positive Acute Angles A and B it is known that sin A = 21/29 and tan B = 28/45. Find the value of cos (A-B) in simplist form.

Answer 1

sinA =21/29

cosA = 20/29

29^2 = 21^2 + 20^2

tanB = 28/45

sinB = 28/52

cosB = 45/53

28^2 + 45^2 = 53^2

cos(A-B) = cosAcosB + sinAsinB

=(20/29)(45/53) + (21/29)(28/53)

= 1488/1537

= about 0.968

check the answer

use a calculator with an inverse trig function

sinA = 21/29, A = 46.4

tanB = 28/45 B=31.9

A and B are positive and acute (between 0 and 90)

A-B = 14.5

cos14.5 = 0.968

Answer 2

does [(√(29²-21²)(21)/29)]+[(28*45)/√(28²+45²)]

look right

cos(A-B)=sinAcosA+sinBcosB

28/45=4/9, so simpler might be

[(√(29²-21²)(21)/29)]+[(4*9)/√(4²+9²)]