Question

Consider the diagram and the derivation below. Given: In △ABC, AD ⊥ BC Derive a formula for the area of △ABC using angle C. It is given that in △ABC, AD ⊥ BC. Using the definition of sine with angle C in △ACD results in sin(C) = . Using the multiplication property of equality to isolate h, the equation becomes bsin(C) = h. Knowing that the formula for the area of a triangle is A = bh is and using the side lengths as shown in the diagram, which expression represents the area of △ABC? bsin(C) absin(C) cbsin(C) hbsin(C) Mark this and return

Answer 1

`A=(1)/(2)a*b*sin(C)`

Explanation:

First consider the diagram:

Now, we know that

`Sin(C)=(AD)/(AC) \\ Sin(C) =(h)/(b) \\h=bsin(C)`

Now, the area of the triangle ABC is given by,

`A=(1)/(2)*BC*AD\\\\A=(1)/(2)*a*h\\\\A=(1)/(2)*a*c*sin(C)\\\\A=(1)/(2)a*b*sin(C)`