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

Answer:

$A=\frac{1}{2}a*b*sin(C)$

Step-by-step explanation:

First consider the diagram:

Now, we know that

$Sin(C)=\frac{AD}{AC} \\ Sin(C) =\frac{h}{b} \\h=bsin(C)$

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

$A=\frac{1}{2}*BC*AD\\\\A=\frac{1}{2}*a*h\\\\A=\frac{1}{2}*a*c*sin(C)\\\\A=\frac{1}{2}a*b*sin(C)$

Answer 2

Answer:

1/2absin(C)

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