Consider the given triangle.
A triangle with vertices labeled A, B, and C is presented. Vertex A is at the top, vertex B is on the lower left, and vertex C is on the lower right. The side opposite vertex A is labeled with the length of 18 units. The angle ABC at vertex B is labeled as 63 degrees, and the angle ACB at vertex C is labeled as 88 degrees, opposite to this angle is side AB labeled with lowercase letter '$c$c'.
First, find the value of $\angle BAC$∠BAC.
Find the length of $c$c.
Round your answer to two decimal places.
Consider the following diagram:
Use the sine rule to prove that the area of $\triangle ABC$△ABC is given by the equation $Area=\frac{a^2\sin B\sin C}{2\sin A}$Area=a2sinBsinC2sinA.
We want to prove that the area of a parallelogram is the product of two adjacent sides and the sine of the included angle.