Find sinA if cosA = ...?

Question

Find sinA if cosA = -3/5 and 90° < A <180°


If possible, could you please show any steps? Thanks in advance to anyone who helps me!

Answer 1

To do this, begin by drawing the picture. Since cosine is negative and the angle is between 90 and 180, the angle is in the second quadrant and the sine will be positive.
Remember that sine is the ratio between opposite and hypotenuse, and cosine is the ratio between adjacent and hypotenuse. So draw a right triangle in the second quadrant, with the acute angle "A" at the origin facing into the second quadrant, like this: _\
Since it faces into the second quadrant, label the side along the x axis -3. and the hypotenuse 5. Now remember the pythagorean relation to find the unknown side. 3^2 + x^2 = 5^2 so x = 4
So the unknown side, opposite to angle A, is 4, so the sine of A is opposite / hypotenuse, that is, 4/5.
For more info see...
http://www.clarku.edu/~djoyce/trig/

Answer 2

The cosine is negative in the 2nd and 3rd quadrant. You are putting it in the 2nd quadrant since 90
So cos A = -3/5 means x = - 3 and r = 5 so y=+4 by Pythagorean theorem
sinA = y/r = 4/5

Answer 3

(sinA)^2+(cosA)^2=1
(sinA)^2+(-3/5)^2=1
(sinA)^2+9/25=1
(sinA)^2=1-9/25
(sinA)^2=16/25
sinA=+/-4/5
since 90° < A <180°,hence
sinA=4/5

Answer 4

As A is in the second quadrant sin A will be positive
Therefore,SinA=+sqrt(1-cos^A)
=+sqrt(1-9/25)
=+sqrt(16/25)
=4/5 ans

Answer 5

Okay, so cosine is x/r and sine is y/r. (tangent is y/r)

Anyway when you say that a is between 90 and 180 it means that your cosine is negative. well, a radius can't be negative, so it's gotta be the x that is. (which makes sense)

so, you know that x= -3 and that r= 5. use the Pythagorean theorem to find y. x2+y2=r2. (-3)^2=y^2=(5)^2

you get 9+y^2=25. now solve for y. you get y^2=16

square root each side and you get y to equal 4. now, since you know that sine is y/r you know that sine is 4/5. (and sine is positive in the first and second quadrants, so you know your answer has to be positive)

Answer 6

Draw the angle. It would be in the second quadrant. adjacent side is three, hypotenuse is 5. By the Pythagorean theorem, the opposite side would be 4. So the sine of A would be 4/5

Answer 7

if cos A = -3/5, it's in quadrant II
because cos A = x/r, where r is the hypothenous (pythagorean theorem)
so, if x=-3, r=5, and A is in quad II, then
r^2 = x^2 + y^2
y^2 = r^2 - x^2
y = 4

sin A = y/r
sin A = 4/5

Answer 8

Angle A is in second quadrant.
In this quadrant draw triangle OPQ where O is the origin.
OP is hypotenuse = 5
OQ is horizontal = 3
PQ is vertical = 4 (by Pythagoras or 3,4,5 triangle)
sin A = 4 / 5

Answer 9

cosA - sinA = sqrt(2)sinA and sinA + cosA =sqrt(2)cosA squaring: 1 - 2sinAcosA = 2sin^2A and 1 + 2sinAcosA = 2cos^2A adding: 2 = 2 so proved