square root 3/2
This problem can be solved using the Sine Rule :a/sin A = b/sin B = c/sin C 10/sin 45 = AB/sin 75 : AB = 10sin 75 ÷ sin 45 = 13.66 units (2dp)
sin(60 degrees) = 0.8660 approx. The exact value is sqrt(3)/2.
1.5
60 degrees
Use the Sine rule. If L is the length of the longer leg, then L/sin(60) = 6/sin(30) So that L = 6*sin(60)/sin(30) = 12*sin(60) = 12*sqrt(3)/2 = 10.39 units.
The sine of 120 degrees can be found using the unit circle. It is equivalent to sin(180° - 60°), which gives sin(120°) = sin(60°). Since sin(60°) is √3/2, the ratio for sin 120 degrees is √3/2.
Tan(60) = Sin(60)/ Cos(60) Sin(60) = sqrt(3)/2 Cos(60) = 1/2 Hence Sin(60) / Cos(60) = [sqrt(3) / 2] / [1/2} => sqrt(3) / 2 X 2/1 sqrt(3) Hence Tan(60) = sqrt(3) = Numerically = 1.732050808....
sin 300 = -sin 60 = -sqrt(3)/2 you can get this because using the unit circle.
5400
Do you mean sin(x)=sqrt(3)/2? IF so, look at at 30/60/90 triangle. We see the sin 60 degrees is square (root of 3)/2
The sine of 60 degrees, denoted as sin(60°), is equal to √3/2. This value arises from the properties of a 30-60-90 triangle, where the ratio of the lengths of the sides opposite the angles is well-defined. Therefore, sin(60°) is approximately 0.866.
sin(30) = sin(90 - 60) = sin(90)*cos(60) - cos(90)*sin(60) = 1*cos(60) - 0*sin(60) = cos(60).
This problem can be solved using the Sine Rule :a/sin A = b/sin B = c/sin C 10/sin 45 = AB/sin 75 : AB = 10sin 75 ÷ sin 45 = 13.66 units (2dp)
sin(60 degrees) = 0.8660 approx. The exact value is sqrt(3)/2.
sin 60 = √(3)/2 or about 0.866 ■
To find the voltage at 90 degrees given an instantaneous voltage of 225 V at a 60-degree angle, we can assume a sinusoidal function for the voltage. The voltage can be represented as ( V(t) = V_m \sin(\theta) ), where ( V_m ) is the maximum voltage. Since you provided the instantaneous voltage at 60 degrees, you can calculate the maximum voltage ( V_m ) using ( V_m = \frac{225}{\sin(60^\circ)} ). Then, to find the voltage at 90 degrees, you would evaluate ( V(t) ) at that angle, resulting in ( V(90^\circ) = V_m \sin(90^\circ) = V_m ).
sin 57 degrees