Question

consider the diagram and derivation. to derive a formula for the area of △abc using angle c.
given: in △abc, ad⊥bc.
it is given that in △abc, ad⊥bc. using the definition of sine with angle c in △acd results in sin(c)=\frac{h}{b}. using the multiplication property of equality to isolate h, the equation becomes b sin(c)=h.
knowing that the formula for the area of a triangle is a = \frac{1}{2}bh and using the side - lengths as shown in the diagram, which expression represents the area of △abc?
\frac{1}{2}ab sin(c)
\frac{1}{2}b sin(c)

Explanation:

Step1: Recall area formula of triangle

The area formula of a triangle is $A=\frac{1}{2}\times base\times height$. In $\triangle ABC$, if we consider the base as $a$ and the height as the length of $AD = h$.

Step2: Express height in terms of given - side and angle

We know from right - triangle $ACD$ that $\sin(C)=\frac{h}{b}$, so $h = b\sin(C)$.

Step3: Substitute height into area formula

Substitute $h = b\sin(C)$ into the area formula $A=\frac{1}{2}\times base\times height$ with base $a$. We get $A=\frac{1}{2}ab\sin(C)$.

Answer:

$\frac{1}{2}ab\sin(C)$