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How do you calculate the area of anisosceles trapezoid if you only know the width of the base and the length and angle of the sides?
One way would be as follows: Let b represent the length of the base, l the length of each of the two sides, and theta the angle between the base and the two sides of length l. Now drop a perpendicular line from each vertex at the top of the trapezoid to the base. This yields two right triangles and a rectangle in the middle. The height of each right triangle (as well as the height of the rectangle) equals l*sin(theta) [because sin(theta)=opposite/hypotenuse] and the length of the base of each right triangle is l*cos(theta). The base of the rectangle is b minus the lengths of the two right triangles. Area of the trapezoid=2*area of each right triangle+area of the rectangle=2*(1/2)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=b*l*sin(theta)-l2*sin(theta)*cos(theta)
What does the Sin of theta equal to if half of the sin of theta. what are the values of theta?
[]=theta 1. sin[]=0.5sin[] Subtract 0.5sin[] from both sides.2. 0.5sin[]=0. Divide both sides by 0.5.3. Sin[] =0.[]=0 or pi (radians)
What is the derivative of sin under root theta?
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
How do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
What is the equivalent of sine theta over secant theta?
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
