We will follow three steps to derive the following formula for the area of a triangle.
Area=1/2absinCThis is the first step. We will start by drawing the altitude from vertex B to AC on the given diagram, and we will label the altitude as h.
Next, let's write a formula for the area of the triangle using the altitude h. Recall that the most common formula for the area of a triangle is one-half of the product of the base and the altitude perpendicular to it.
Area=1/2 bh
b
We will follow three steps to derive the following formula for the area of a triangle.
Area=1/2absinC
This is the second step. We are asked to write an equation for sinC. Let's consider the diagram from Part A.
We can see that ∠C is an acute angle of a right triangle △ BCD. The sine of ∠C is the ratio of the length of the leg opposite ∠C and the length of the hypotenuse.
sinC=Length of leg opposite∠C/Length of hypotenuse
Note that the length of the leg opposite ∠C is h, and the length of the hypotenuse is a.
sinC=h/a
We obtained an equation for sinC.
c
We will follow three steps to derive the following formula for the area of a triangle.
Area=1/2absinC
This is the last step. We will use the results from Part A and Part B to write a formula for the area of a triangle that does not include h. Let's recall the results from Part A and Part B.
& Part A & & Part B [0.8em]
&Area=1/2bh & &sinC=h/a
Our goal is to eliminate h from the formula from Part A. To do it, we will express h in terms of sinC and a by solving the equation from Part B for h.
