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How do you find tan-theta when sin-theta equals -0.5736 and cos-theta is greater than 0?
-0.5736
What is a quadrantal angle?
A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.
If the resultant of two vectors each of magnitide f is twice of magnitude F then angle bw?
If the resultant of two vectors, each of magnitude ( f ), is twice the magnitude ( F ), then the angle ( \theta ) between the two vectors can be determined using the formula for the resultant of two vectors: [ R = \sqrt{f^2 + f^2 + 2f^2 \cos \theta} ] Given that ( R = 2F ), we set ( R = 2f ) (assuming ( F = f )). This leads to the equation ( 4f^2 = 2f^2(1 + \cos \theta) ). Solving for ( \theta ), we find that ( \cos \theta = 0 ), which means ( \theta = 90^\circ ). Thus, the angle between the vectors is ( 90^\circ ).
What is the angle of elevation of the sun when a flagpole M tall cast a shadow m long?
The angle of elevation of the sun can be determined using the tangent function in trigonometry. Specifically, if the height of the flagpole is ( M ) and the length of the shadow is ( m ), the angle of elevation ( \theta ) can be calculated using the formula ( \tan(\theta) = \frac{M}{m} ). To find the angle, use ( \theta = \arctan\left(\frac{M}{m}\right) ). This angle represents how high the sun is in the sky relative to the horizontal ground.
What is the angle of the elevation of the top of a tree 12.7 m from a point 23.7 m away on level ground?
To find the angle of elevation of the top of the tree, we can use the tangent function in trigonometry. The formula is ( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} ), where the opposite side is the height of the tree (12.7 m) and the adjacent side is the distance from the tree (23.7 m). Therefore, ( \theta = \tan^{-1}\left(\frac{12.7}{23.7}\right) ). Calculating this gives approximately ( \theta \approx 28.6^\circ ).
If costheta equals frac2sqrt6 in Quadrant 1 then find tantheta?
tan theta = sqrt(2)/2 = 1/sqrt(2).
If tan Theta equals 2 with Theta in Quadrant 3 find cot Theta?
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
How do you find tan-theta when sin-theta equals -0.5736 and cos-theta is greater than 0?
-0.5736
How do you find the theta of a triangle?
The answer depends on what theta represents!
How do you find perimeter of quadrant?
perimeter of what quadrant?
Should we find cos theta if sec theta equals -10?
No.
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
An angle is 12 degrees more than its supplement Find the angle?
96 degrees Let theta represent the measure of the angle we are trying to find and theta' represent the measure of its supplement. From the problem, we know: theta=theta'+12 Because supplementary angles sum to 180 degrees, we also know: theta+theta'=180 Substituting the value from theta in the first equation into the second, we get: (theta'+12)+theta'=180 2*theta'+12=180 2*theta'=180-12=168 theta'=168/2=84 Substituting this value for theta' back into the first equation, we get: theta+84=180 theta=180-84=96
Find all angles in the interval 0 360 satisfying the equation cos theta equals 0.7902?
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
What quadrant would you find pi over 5 in?
Pi / 5 would be in Quadrant I.
How can you find the length of an arc with the radius known?
The length of the arc is r*theta where r is the radius and theta the angle subtended by the arc at the centre of the circle. If you do not know theta (or cannot derive it), you cannot find the length of the arc.
How do you find the sine of an angle?
sine[theta]=opposite/hypotenuse=square root of (1-[cos[theta]]^2)
