Asked by Kealeboga
Given that cosA=4/5 and sinB=15/17 where A and B are acute angles,find the value of cos(A-B)
Answers
Answered by
Reiny
sketch right-angled triangles that have the given information shown,
both in quadrant I
cosA = 4/5 = x/r --->x=4 and r=5
x^2 + y^2 = r^2
16 + y^2 = 25
y = √9 = 3
then sinA = 3/5
sinB=15/17 = y/r, so y=15, r = 17
x^2 + 15^2 = 17^2
x = √64 = 8
cosB = 8/17
You should know that
cos(A-B) = cosAcosB + sinAsinB
= ... , you know all the values
(4/5)(8/17) + 3/5(15/17) =
both in quadrant I
cosA = 4/5 = x/r --->x=4 and r=5
x^2 + y^2 = r^2
16 + y^2 = 25
y = √9 = 3
then sinA = 3/5
sinB=15/17 = y/r, so y=15, r = 17
x^2 + 15^2 = 17^2
x = √64 = 8
cosB = 8/17
You should know that
cos(A-B) = cosAcosB + sinAsinB
= ... , you know all the values
(4/5)(8/17) + 3/5(15/17) =
Answered by
oobleck
or, if you just recognize a few basic Pythagorean triples, like 3-4-5, 5-12-13, and 8-15-17 you could have skipped directly to the last step.
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