Show that twice the area of triangle ABC=p(m+n)sin thita
To show that twice the area of triangle ABC is equal to p(m+n)sin(theta), we can use the formula for the area of a triangle.
The formula for the area of a triangle using the lengths of two sides and the included angle is given by:
Area = 0.5 * a * b * sin(theta),
where "a" and "b" are the lengths of two sides of the triangle, and "theta" is the included angle between those two sides.
In this case, we want to show that twice the area of triangle ABC is equal to p(m+n)sin(theta).
Let's assume that side AB has a length of "m" and side AC has a length of "n".
Since we are given that twice the area of triangle ABC is p(m+n)sin(theta), we can denote it as:
2 * Area = p(m+n)sin(theta).
Now, we can use the formula for the area of a triangle to calculate the area of triangle ABC:
Area = 0.5 * AB * AC * sin(theta).
Substituting the values, we get:
Area = 0.5 * m * n * sin(theta).
Multiplying both sides of the equation by 2, we have:
2 * Area = 2 * (0.5 * m * n * sin(theta)).
Simplifying this expression gives:
2 * Area = m * n * sin(theta).
Comparing this with the given expression, we can see that:
2 * Area = p(m+n)sin(theta).
Hence, we have shown that twice the area of triangle ABC is equal to p(m+n)sin(theta).
To show that twice the area of triangle ABC is equal to p(m+n)sin(theta), we can use the formula for the area of a triangle and the definition of sine:
1. The formula for the area of a triangle is given by A = 0.5 * base * height, where base is the length of any side of the triangle and height is the perpendicular distance from the base to the opposite vertex.
2. Let's assume that side AB of triangle ABC is the base. The height of the triangle can be represented as h.
3. To calculate h, we can use the sine of angle theta. Recall that the sine of an angle in a right triangle is equal to the ratio of the length of the opposite side to the length of the hypotenuse. In triangle ABC, the side opposite angle theta is BC, and the hypotenuse is AC.
Therefore, sin(theta) = (BC / AC)
4. Rearranging this equation, we have BC = (AC * sin(theta)).
5. Now, substituting BC into the formula for the area, we have A = 0.5 * AB * (AC * sin(theta)).
6. Since we want to find twice the area of triangle ABC, we multiply both sides of the equation by 2:
2A = AB * (AC * sin(theta)).
7. Finally, we can see that the right-hand side of the equation is equal to p(m+n)sin(theta), where p is a constant and (m+n) is the product of AC and AB. Therefore:
twice the area of triangle ABC = p(m+n)sin(theta).