How do you solve this trig. problem?

Question

A storm drain has a cross section in the shape of an isosceles trapezoid. The shorter base and each of the two equal sides measure 2 m, and x is the angle formed by the longer base of each of the equal sides. Write an expression for the area, A, of the cross section in terms of sin x and cos x.

Answer 1

area = 1/2(b1+b2)h

b1 = 2
b2 = 2 + 2cosx + 2cosx
h = 2sinx

area = 1/2(4+4cosx)2sinx
=4(1+cosx)sinx

Answer 2

you ought to memorize what applications are advantageous wherein quadrants and that are unfavorable. which will help do those issues directly. in spite of the fact that that is not too complicated to make certain in case you comedian strip the 4 quadrants and bear in mind that the radius (the hypotenuse) is often advantageous. interior the 1st query, you're given the sine, so which you comprehend 2 factors of a triangle and you could locate the 0.33 ingredient by way of the tPythagorean theorem. that is 3(sqrt 5). you could now fill in different functionality values, remembering that any functionality that is composed of the adjoining ingredient (which suits to a unfavorable x fee) would be unfavorable. interior the 2d query, the sine is unfavorable, so which you comprehend the attitude must be interior the 0.33 or fourth quadrant. The cotangent is advantageous, so the tangent is advantageous too (the cotangent is the reciprocal of the tangent). The cotangent is unfavorable interior the fourth quadrant, so the attitude won't be able to be there, so it could be interior the 0.33 quadrant. The cotangent is 4/3, and that's the adjoining over the different, so the hypotenuse ought to be 5 (a three-4-5 good triangle). you could now comedian strip the attitude. interior the 0.33 quadrant, the two the x and y values are unfavorable, so which you would be able to fill interior the sign in keeping with whether the fraction -- the ratio that defines the functionality -- has 2 unfavorable numbers or one advantageous and one unfavorable.

Answer 3

Area of Each triangle =1/2 Base x Height
A(t) = 1/2(1/2(Big Base - Small Base) x (2sin(X)))
A(t) = 1/2(1/2((2 + 4cos(X)) - 2) x (2sin(X)))
A(t) = 1/2(2 + 2cos(X) - 2) x sin(X)
A(t) = 1/2(2cos(X) x sin(X))
A(t) = cos(X) x sin(X)

Area of mid section = Length x Width
A(m) = (2 x 2sin(X)) x 2
A(m) = 8sin(X)

Total Area = Area of each triangle + Area of Midsection
8sin(X) + 2(cos(X) x (sin(X))
(4sin(X))(2+cos(X)