On the unit circle sin(90) degrees is at Y = 1 and as that is on the Y axis X will equal = 0.
Ask yourself. Where would 90 degrees be on a 360 degree circle? Straight up.
A right angle is 90 degrees.
The Unit Circle is a circle that has a radius of 1 and a center at the origin. If you look at the unit circle 90 degrees is at the point (0,1). Cosine is equal to the x value of a point on the Unit Circle. The line created to the point (0,1) on the unit circle when the degree is 90 is completely vertical, which in turn makes the x value 0 and thus, cosine of 90 = 0. Read more >> Options >> http://www.answers.com?initiator=FFANS
tan0.15
150 degrees
Greater than 90 degrees but less than 180 degrees.
sin(90) = 1
cos(35)sin(55)+sin(35)cos(55) If we rewrite this switching the first and second terms we get: sin(35)cos(55)+cos(35)sin(55) which is a more common form of the sin sum and difference formulas. Thus this is equal to sin(90) and sin(90)=1
a central angle does NOT have to equal 90 degrees
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
Two angles that are equal to 90 degrees are Complimentary.
Angles combining to equal 90 degrees are known as complementary angles.
No because they equal 180 degrees
If all 4 sides are equal and the angles are not 90 degrees it is a rhombus.
90 degrees Fahrenheit = 32.2 degrees Celsius.
90 degrees Fahrenheit is 32.2 (repeating) degrees Celsius.
On the unit circle at 90 degrees the 90 degrees in radians is pi/2 and the coordinates for this are: (0,1). The tan function = sin/cos. In the coordinate system x is cos and y is sin. Therefore (0,1) ; cos=0, & sin=1 . Tan=sin/cos so tan of 90 degrees = 1/0. The answer of tan(90) = undefined. There can not be a 0 in the denominator, because you can't devide by something with no quantity. Something with no quantity is 0. Or, on a limits point of view, it would be infinity.
The temperature 90 degrees Celsius is equal to 194 degrees Fahrenheit. Which is to say very hot.