Question

Show that for any triangle △ABC the area of the triangle [ABC]=1/2AC * BCsinC. How do i solve this?

Answer 1

Substituting in the "classic" formula `[ABC] = 1/2*h*b`.

Explanation:

So, the formula for the area of a triangle is `1/2*h*b`.
Now, let there be a point `"D"` on the side `"BC"`, such that `"AD"` is perpendicular to `"BC"`. (Basically, `"AD"` is the height of the triangle and `"BC"` is the base of the triangle.)
Now, in the right angled triangle `ACD`, `sin C = "AD"/"AC"`, so `"AD" = "AC"*sin C`.

Substituting `"AD"` and `"BC"` in the area formula, it gives out that
`[ABC] = 1/2*"AD"*"BC"`

`[ABC] = 1/2*(AC*sin C)*"BC"`

`[ABC] = 1/2AC * BC*sin C`