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Question

Consider the diagram and the derivation below.

Given: In ABC\triangle ABC, ADBC\overline{AD} \perp \overline{BC}

Derive a formula for the area of ABC\triangle ABC using angle CC.

it is given that in ABC\triangle ABC, ABBC\overline{AB} \perp \overline{BC}. Using the definition of sine with angle CC in ACD\triangle ACD results in sinC=hb\sin{C}=\frac{h}{b}. Using the multiplication property of equality to isolate hh, the equation becomes bsin(C)=hb\sin(C)=h.

Knowing that the formula for the area of a triangle is A=12bhA=\frac{1}{2}bh and using the side lengths as shown in the diagram, which expression represents the area of ABC\triangle ABC?

\bullet 12bsin(C)\frac{1}{2}b\sin(C)

\bullet 12absin(C)\frac{1}{2}ab\sin(C)

\bullet 12cbsin(C)\frac{1}{2}cb\sin(C)

\bullet 12hbsin(C)\frac{1}{2}hb\sin(C)'slader'

Solution

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Substitute h=bsinCh=b\sin C to the area formula of the triangle where b=ab=a (base of the triangle with corresponding height hh)

A=12bhA=\dfrac{1}{2}bh

A=12a(bsinC)A=\dfrac{1}{2}a(b\sin C)

A=12absinCA=\color{#c34632}\dfrac{1}{2}ab \sin C

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