If tanB= -6/8 and sinB<0, find cosB?

Answer 1

Two ways to do this:

(1) Use inverse tangent, then just take the cosine:

If tanB = -6/8,
B = -36.87 degrees

cosB = 0.8

(2) Use pythagoras:

Tangent represents the ratio of the opposite (6) and adjacent (8) legs of a right triangle. If the legs of a right triangle are 6 and 8, the hypotenuse is 10 (it's a 3:4:5 right triangle scaled up by a factor of 2, or sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10).

Thus...
The sin = opposite/hypotenuse = 6/10
The cos = adjacent/hypotenuse = 8/10

When doing it this way, you lose the sign of the values, though, so you have to remember that the tangent is sine divided by cosine. When the tangent and sine are both negative, the cosine is positive.

Answer 2

Ignore the sign for a moment. Draw a right angled triangle where one angle would have a tan of 6/8 ie perpendicular=6 and base=8. Calculate the hypotenuse using Pythagoras Theorem to get 10 (6^2 + 8^2 = 10^2).

|tanB|=6/8 (perpendicular/base)
|cosB|=8/10 or 0.8
tanB=sinB/cosB
tanB=-6/8<0
sinB<0
Therefore cosB>0 (negative number/positive number = negative number)
cosB=0.8

Answer 3

okay. by saying tan and sin are negative, they imply the triangle is in quad. III (meaning tan and sin are neg., but cos is positive.). If tan B= -6/8 (opposite/adjacent).

cosB =8/10 (adjacent/hypotenuse)

(to find hypoteneuse)
(opposite^2+adjacent^2=hypotenuse^2, which is pythagorean's thyrum.(<- SP?))
6^2+8^2=c^2
36+64=c^2
100=c^2
10=c