Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 298: 39

$\frac{\sqrt 3}{2}$ sq. units

We first use the formula $A=\frac{1}{2}bh$ to find the area of the triangle: $A=\frac{1}{2}bh$ $A=\frac{1}{2}(1)(\sqrt 3)$ $A=\frac{\sqrt 3}{2}$ Now, we use the formula $Area=\frac{1}{2}ab \sin C$ to verify the result: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(1)(\sqrt 3) \sin 90$ $Area=\frac{\sqrt 3}{2} \sin 90$ $Area=\frac{\sqrt 3}{2}(1)$ $Area=\frac{\sqrt 3}{2}$ Therefore, both the formulas give the same area which is $\frac{\sqrt 3}{2}$ sq. units.