An earlier section,
Area Formulas: Triangle,
Parallelogram, Trapezoid,
showed how to find the area of a
triangle with a base/height pair.
The key results
are summarized here for convenience:

Drop a perpendicular to the opposite side ... 
... or, to an extension of the opposite side. 
A derivation of the triangle area formula ($\,bh/2\,$) was given in Area Formulas: Triangle, Parallelogram, Trapezoid for the situation where the altitude ‘hits’ the base (leftmost picture above). The formula also works when the altitude doesn't ‘hit’ the base, but instead hits an extension of the base, as shown at right. Here's how:

area of blue triangle is $\,\frac 12 bh\,$ 
Note that there are infinitely many triangles
with a given base/height pair.
For example, all the triangles below have the same base/height pair.
Same base/height pair—same area!
A base/height pair uniquely determines the area of a triangle, but
not the shape of the triangle.
All these triangles have the same base/height pair, hence the same area. 
By the
SAS (sideangleside) triangle congruence theorem, two sides and an included angle uniquely defines a triangle. Let $\,a\,$ and $\,b\,$ denote two sides of a triangle, with included angle $\,\theta\,$. The sketches at right and below show the two situations that can occur when finding the altitude that has corresponding base $\,b\,$. In both cases, the altitude length is found using the yellow triangle (which is a right triangle with hypotenuse $\,a\,$). In both cases, the altitude length is $\,h = a\sin\theta\,$. Thus: $$ \begin{align} &\text{area of triangle with sides $a$ and $b$ and included angle $\theta$}\cr\cr &\qquad = \frac 12(\text{base})(\text{height})\cr\cr &\qquad = \frac 12 b(a\sin\theta)\cr\cr &\qquad = \frac 12 ab\sin\theta \end{align} $$ In words, the area is: times the sine of the included angle 

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
