Stage 1 - Stage 3

Triangle Problems

Lesson

So far we have found unknown side lengths using Pythagoras' theorem and then looked at 3 special ratios that we can use to find unknown sides or angles in right-angled triangles.

Right-angled triangles

Pythagoras' theorem: $a^2+b^2=c^2$a2+b2=c2, where c is the hypotenuse

$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse = $\frac{O}{H}$OH

$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse = $\frac{A}{H}$AH

$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent =$\frac{O}{A}$OA

Problem solving in trigonometry can be in finding unknowns like we have already been doing, using trigonometry in real world applications or in solving geometrical problems like these.

Examples

Question 1

Find $x$x in the following geometrical diagram,

Think: In order to find $x$x, I will need to identify some other measurements along the way. My problem solving strategy will be

1. Find length $AC$AC using trig ratio sine

2. Find length $ED$ED, $\frac{AC}{3}$AC3

3. Find length $x$x, using trig ratio sine

Do:

1. Find length $AC$AC using trig ratio sine

$\sin23^\circ=\frac{43.6}{AC}$sin23°=43.6AC

$AC=\frac{43.6}{\sin23^\circ}$AC=43.6sin23°

$AC=111.59$AC=111.59

2. Find length $ED$ED, $\frac{AC}{3}$AC3

$ED=\frac{111.59}{3}$ED=111.593

$ED=37.2$ED=37.2

3. Find length $x$x, using trig ratio sine

$\sin35.6^\circ=\frac{x}{37.2}$sin35.6°=x37.2

$x=37.2\times\sin35.6^\circ$x=37.2×sin35.6°

$x=21.65$x=21.65

Question 2

Find the length of the unknown side in this right-angled triangle, expressing your answer as a decimal approximation to two decimal places.

Question 3

Find the value of $f$f, correct to two decimal places.

Question 4

Find the value of $x$x to the nearest degree.

Question 5

Find the value of $x$x, the side length of the parallelogram, to the nearest centimetre.