IF cosx°=3/5 find the exact value of sinx° for 0°<x°<90°?

Answer 1

This means X is an angle in a 3/4/5 right triangle, so sinx will be 4/5.

Answer 2

Sin x =4/5 because as you know

sin^2 +coc^2=1 so sinx=4/5 or -4/5 but because 0

Answer 3

arccos(3/5)=53.13
sin(53.13)=.8
But w/o a calculator, use Pythagorean Theorem
and compute the oposite side.
5^2-3^2=25-9=16=x^2
x=4
So the sinx is 4/5 OR
use the trig identity of
sinx^2 + cosx^2 =1 which happens to be derived from
the Pythagorean Theorem and we have
1-(3/5)^2 = sinx^2
1-9/25=16/25
sinx=4/5


OK

Answer 4

Simple. You are talking of a right angled tiangle
with sides3,4,5
sin x=(1-cos^2x)^1/2
=(1-9/25)^1/2=(16/25)^1/2=4/5

Answer 5

sinx=(1-cos^2x)^1/2
=(1-9/25)^1/2=(16/25)^1/2=4/5
the negative value is ignored as
0*

Answer 6

cos x = 3/5
squarring both side
(cos x)^2=9/25
1 - (sin x)^2 = (9/25)
(sin x)^2 = 1- (9/25)
=16/25
Taking sqrt
sin x = 4/5 or -4/5
Since in first qurant sin x is positive therefore reject negative value
Hence sin x =4/5=0.8

Answer 7

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Answer 8

sin x = sqrt(1 - cos^2x) = sqrt(1 - 9/25) = 4/5,
since we are in the first quadrant.