A+ » VCE » Further Maths U3 & 4 Master Notes » OA3 Geometry and Measurement » 1.6 Solving Triangles in 2 and 3 dimensions

1.6 Solving Triangles in 2 and 3 dimensions

Note: if you cannot remember the trigonometric identities, revise notes for 1.5 Methods for Solving Triangles.

Solving for Area without Height

  • In some cases, triangles will be given with the edge lengths rather than the height. In this case our regular formula for the area of a triangle (A=\frac{1}{2} b h) needs some reworking. If we draw a line down the middle of a triangle, we can effectively split it into two right-angled triangles from which we can use our trigonometric identities to solve for the height. Using our formula for sine, we find:

\sin (A) =\frac{h}{c}

h =c \sin (\theta)

Consequently:

A=\frac{1}{2} b c \sin (\theta)

Picture 8
  • This formula works for non-right angled triangles as well.

Note: trigonometry is explored in more detail in notes 2.3 Solving Triangles using Trigonometry and 2.4 Applications of Trigonometry and Pythagoras Theorem.

This content is for Master Notes FM members only. Unlock the content by signing up for a membership level - quick and easy!
Log InSign Up