Math Geometry Sine, Cosine and Tangent Sine, Cosine and Tangent on a right triangle Exercises: Sine, Cosine and Tangent on a right triangle

Exercises: Sine, Cosine and Tangent on a right triangle

This sketch, which is not to scale, shows a rectangle with side lengths $a = 5.0 cm$ and $b = 7.0 cm$ .

Calculate the angle $α$ .

For this task you need the following basic knowledge: Sine, Cosine and Tangent

First, draw a sketch.

Draw a rectangle and point out the angle $α$ .

If you now draw a line from the intersection of the diagonals to the centre of the side $[B C]$ , then the angle $α$ you are looking for is divided into two equal parts by this auxiliary line.

Skizze Rechteck mit eingezeichneter Hilfslinie

There are several ways to calculate the angle $\frac{α}{2}$ :

You may compute $\frac{α}{2}$ using the sine.
You may compute $\frac{α}{2}$ using the tangent.

The solution with the tangent is mathematically easier .

Solution with the sine

First we calculate the distance from $A$ to $C$ and thus obtain a right-angled triangle.

\begin{array}{rcl} A C & = & \sqrt{b^{2} + a^{2}} \\ A C & = & \sqrt{(7 cm)^{2} + (5 cm)^{2}} \\ A C & = & \sqrt{74} cm \end{array}

To calculate the angle $α$ , we divide the distance from $A$ to $C$ and the distance from $A$ to $B$ by $2$ . Thus we can apply the sine to calculate $\frac{α}{2}$ . Then we multiply the calculated angle by $2$ and get the angle we are looking for, $α$ .

\frac{A B}{2} = \frac{5 cm}{2} = 2.5 cm \frac{A C}{2} = \frac{\sqrt{74}}{2} cm \sin (\frac{α}{2}) = \frac{\frac{A B}{2}}{\frac{A C}{2}} = \frac{2.5}{\frac{\sqrt{74}}{2}} α = 2 \cdot \sin^{- 1} (\frac{2.5}{\frac{\sqrt{74}}{2}}) \approx 71 °

Solution with the tangent

Consider the triangle $Δ M M_{B C} C$ . This triangle is right-angled.

The side $[M M_{B C}]$ is the adjacent cathetus of $\frac{α}{2}$ .
The side $[M_{B C} C]$ is the opposite cathetus of $\frac{α}{2}$ .

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