When considering right-angled triangles we need to know which side is the hypotenuse (the side opposite the right angle). Then the sides which we label as opposite and adjacent sides are always in relation to a particular angle.
Finding side lengths
If we know $2$2 sides and want to find the third, we would use Pythagorean formula.
$a^2+b^2=c^2$a2+b2=c2
If we know $1$1side length and an angle, we would use one of the trigonometric ratios.
The most common mistake is when the wrong ratio is used. We have to remember the ratios and the sides that apply to those ratios. For most students the mnemonic SOHCAHTOA can be a great help.
$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$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
Practice questions
Question 1
Find the value of $f$f, correct to two decimal places.
Question 2
Find the value of $f$f, correct to two decimal places.
A right-angled triangle with a 29-degree opposite to its base is shown. The side adjacent the 29-degree angle is labeled as "$f$f cm" indicating a length in centimeters that is not specified. The hypotenuse is labeled "7 cm," indicating a length of 7 centimeters.
The side opposite to the 29-degree angle is not labeled. The right angle is indicated by a small square at the triangle's base
Question 3
Find the length $g$g, correct to two decimal places.
Finding the size of an unknown angle
We can also use the trigonometric ratios to find the size of unknown angles. To do this we need any $2$2of the side lengths.
We identify which ratio we need to use, we write the rule, we fill in the information which we know then use inverse operations to rearrange and solve.
Practice questions
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.
Problem solving with right-angled triangles
We can use trigonometry in real world applications or in solving geometrical problems. We also come across the angle of elevation and depression in applications of right-angled triangles.
| Angle of Elevation: the angle made between the line of sight of the observer and the 'horizontal' when the object is ABOVE the horizontal (observer is looking UP) |  |
| Angle of Depression: the angle made between the line of sight of the observer and the 'horizontal' when the object is BELOW the horizontal (observer is looking DOWN) |  |
Practice questions
Question 5
A ladder measuring $2.36$2.36 m in length is leaning against a wall. If the angle the ladder makes with the ground is $y$y$^\circ$°, find $y$y to two decimal places.
Question 6
Consider the given figure.
Find the unknown angle $x$x, correct to two decimal places.
Find $y$y, correct to two decimal places.
Find $z$z correct to two decimal places.
Question 7
Georgia is riding her pushbike up a hill that has an incline of $7^\circ$7°.
If she rides her bike $689$689 metres up the hill, what is $d$d, the horizontal distance from where she started, to the nearest metre?
question 8
In the following diagram, $\angle CAE=61^\circ$∠CAE=61°, $\angle CBE=73^\circ$∠CBE=73° and $CE=25$CE=25.
Find the length of $AE$AE, correct to four decimal places.
Find the length of $BE$BE, correct to four decimal places.
Hence, find the length of $AB$AB, correct to two decimal places.
Find the length of $BD$BD, correct to one decimal place.