Key Stage 4

Solve for sides or angles II

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.

A right-angled triangle with an interior angle of $25$25 degrees. The side adjacent to the $25$25-degree angle has a length of $11$11 mm and its opposite side measures $f$ mm.

Question 4

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

A right-angled triangle with vertices labeled $A$ , $B$ and $C$ . Vertex $A$ is at the top, $B$ at the bottom right, and $C$ at the bottom left. A small square at vertex $A$ indicates that it is a right angle. Side $interval(BC)$ , which is the side opposite vertex $A$ , is the hypotenuse and is marked with a length of $25$ . The angle located at vertex B is labelled $x$ . Side $interval(AB)$ , descending from the right angle at vertex $A$ to vertex $B$ , is marked with a length of $7$ , and is adjacent to the angle $x$ . Side $interval(AC)$ is opposite the angle $x$ .

Question 5

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

A parallelogram with its angle at upper-right corner labeled as $52$ degrees, indicating its measure. Other angles of the parallelogram are not labeled. Its top side is labeled $32$ cm, indicating its length. Its left side is labeled $x$`x` cm, indicating its unknown length. A perpendicular internal segment is drawn from the upper-left corner to the bottom side of the parallelogram, thus creating a right-angled triangle on the left side. The bottom side of the parallelogram, which is parallel to the top side as indicated by the double arrowheads, is cut into two parts: the left part with no measurement indicated and the right part which measures $11$ cm, as labeled.

In the triangle, the sides are the internal segment, the left side of the parallelogram, and the left part of the bottom side of the parallelogram. A square symbol is shown at the corner where the internal segment and the bottom side intersect to indicate that it is a right angle. The left side of the parallelogram labeled $x$`x` cm acts as the hypotenuse of the triangle. The internal segment is the vertical leg of the triangle. The left part of the bottom side of the parallelogram acts as horizontal leg of the triangle.