Two sides of a triangle are x units and y units in length. The angle between them is theta. Prove that the area of the triangle is 1/2xysin(theta).

I'm not sure how to do this question, I tried drawing a picture but I'm still not understanding it.

draw a perpendicular from side x to side y, let it be h

Area of triangle
= (1/2)base x height
= (1/2) y h

but sinØ = h/x
h = xsinØ

so area =(1/2) y xsinØ
= (1/2)xy sinØ

To prove that the area of the triangle is equal to 1/2xy sin(theta), we can use the formula for the area of a triangle, which is A = (1/2)base*height.

Let's start by drawing a triangle. Label the two sides of the triangle as x units and y units, and let the angle between them be theta.

Now, we need to find the height of the triangle. To do this, we can drop a perpendicular from one of the vertices to the opposite side. Let's say we drop the perpendicular from the vertex opposite to the side with length x. This will divide the opposite side into two segments, let's call them a and b.

Now, we have a right triangle formed, with the hypotenuse of length y, one leg of length a, and the angle between them is theta.

Using trigonometry, we can express a in terms of x and theta. Since a is the adjacent side to the angle theta, we can use the cosine function: cos(theta) = a/y. Solving for a, we get a = y*cos(theta).

Now, to find b, we can use the Pythagorean theorem: b^2 + a^2 = x^2. Substituting the value of a we found earlier, we have b^2 + (y*cos(theta))^2 = x^2. Simplifying, we get b^2 = x^2 - y^2*cos^2(theta). Taking the square root of both sides, we get b = sqrt(x^2 - y^2*cos^2(theta)).

Now, we have the base of the triangle as x, and the height of the triangle as b. Substituting these values into the formula for the area of a triangle, we get:

A = (1/2) * x * b
= (1/2) * x * sqrt(x^2 - y^2*cos^2(theta)).

At this point, we can simplify the expression further. Recall that sin^2(theta) + cos^2(theta) = 1. Rearranging this equation, we get cos^2(theta) = 1 - sin^2(theta). Substituting this into our expression, we have:

A = (1/2) * x * sqrt(x^2 - y^2*(1 - sin^2(theta)))
= (1/2) * x * sqrt(x^2 - y^2 + y^2*sin^2(theta))
= (1/2) * x * sqrt(x^2 - y^2 + (y*sin(theta))^2)
= (1/2) * x * sqrt((x^2 - y^2)*(x^2 - y^2 + (y*sin(theta))^2)).

We can further simplify this expression by noticing that (x^2 - y^2) is a difference of squares, and it can be factored as (x+y)(x-y). Substituting this factorization into our expression, we get:

A = (1/2) * x * sqrt((x+y)(x-y)*(x^2 - y^2 + (y*sin(theta))^2))
= (1/2) * x * sqrt((x+y)(x-y))*(sqrt(x^2 - y^2 + (y*sin(theta))^2)).

Recall that sqrt(x^2 - y^2 + (y*sin(theta))^2) is the length of b, which is the other leg in the right triangle we formed. So, we can substitute b into our expression:

A = (1/2) * x * sqrt((x+y)(x-y))*b
= (1/2) * x * sqrt((x+y)(x-y))*sqrt(x^2 - y^2*cos^2(theta))
= (1/2) * x * sqrt((x+y)(x-y)(x^2 - y^2*cos^2(theta))).

Finally, we can notice that (x+y)(x-y) is another difference of squares, which can be factored as (x^2 - y^2). Substituting this factorization into our expression, we get:

A = (1/2) * x * sqrt(x^2 - y^2)*sqrt(x^2 - y^2*cos^2(theta))
= (1/2) * x * (x^2 - y^2)*sin(theta)
= (1/2) * x * y * sin(theta).

Therefore, we have proven that the area of the triangle is equal to 1/2xy sin(theta).