Right-angled Triangles

Right-angled triangles is the Standard 1 measurement topic for Year 12. Every question has a right angle somewhere, often hidden in a practical picture: a ramp, a ladder, a cliff, a walk on a bearing.

• Pythagoras' theorem to find a side, in practical two-dimensional problems.
• Trigonometric ratios (sine, cosine, tangent) to find an unknown side.
• Inverse ratios to find an unknown angle, in degrees and in degrees and minutes.
• Angles of elevation and depression.
• Compass bearings (like N30°E) and true bearings (like 030°), and converting between them.

The sine rule, cosine rule and non-right-angled triangles are Standard 2 only.

In a right-angled triangle, the longest side, opposite the right angle, is the hypotenuse $c$:

• Finding the hypotenuse: add the squares, then take the square root.
• Finding a shorter side: subtract from the square of the hypotenuse, then take the square root.

Pythagoras' theorem needs two known sides. If you know a side and an angle, use trigonometry instead.

Label the sides from the angle you are using, $\theta$:

• Hypotenuse (H): the longest side, opposite the right angle.
• Opposite (O): the side across from $\theta$.
• Adjacent (A): the side next to $\theta$ that isn't the hypotenuse.

A memory aid is SOH CAH TOA. Before you start, check your calculator is in degrees mode (a "D" on the screen).

Finding a side

Choose the ratio that uses the side you know and the side you want. If the unknown is on top of the fraction, multiply; if it's on the bottom, divide.

Worked example. A support wire is attached to a pole 9.5 m above the ground and makes an angle of 62° with the level ground. How long is the wire?

The 9.5 m is opposite the 62° angle and the wire is the hypotenuse, so use sine.

Check: the hypotenuse must be longer than the other sides, and 10.76 > 9.5.

Finding an angle

Use the inverse function: ${\sin }^{-1}$, ${\cos }^{-1}$ or ${\tan }^{-1}$ (usually SHIFT then the ratio key).

Worked example. A ramp rises 0.8 m over a horizontal distance of 9.6 m. Find the angle the ramp makes with the ground, to the nearest minute.

Opposite and adjacent are known, so use tangent.

Degrees and minutes

One degree is 60 minutes: $1^{\circ }=60^{\prime }$. To convert the decimal part, multiply by 60:

Most calculators do this with the degrees–minutes–seconds key (° ′ ″). To enter an angle like 38°20′, use the same key, or convert: $20^{\prime }={\frac{20}{60}}^{\circ }$ so 38°20′ is $38.{\dot{3}}^{\circ }$. A common error is typing 38.20, which is 38°12′, not 38°20′.

Both are measured from the horizontal.

• Angle of elevation: looking up from the horizontal to an object.
• Angle of depression: looking down from the horizontal to an object.

The angle of elevation from A to B equals the angle of depression from B to A, because the two horizontal lines are parallel (they are alternate angles). This lets you put the angle at the bottom of the triangle, where it is easier to label.

Worked example. A drone hovers 120 m above flat ground. The angle of depression from the drone to a parked car is 23°. How far is the car horizontally from the point directly below the drone?

The angle of elevation from the car to the drone is also 23°. From the car, 120 m is opposite and the horizontal distance $d$ is adjacent.

Draw a diagram for every elevation or depression question: a horizontal line, the line of sight, and the angle between them.

A bearing gives a direction.

• Compass bearing: start at N or S, then turn a number of degrees towards E or W. For example, S40°E means face south and turn 40° towards east.
• True bearing: measured clockwise from north, written with three digits. For example, 040°, 135°, 300°.

The reverse bearing (back the other way) differs by 180°: if B is on a bearing of 070° from A, then A is on a bearing of 250° from B.

Worked example: a walk on two bearings

A hiker walks 3 km due east from camp C to point P, then 4 km due south to point Q. Find the distance and the true bearing of Q from C.

Due east and due south are perpendicular, so triangle CPQ has a right angle at P.

The angle at C, measured from east down towards south:

East is 090°, and Q is a further 53.13° clockwise, so the true bearing of Q from C is about $90^{\circ }+53^{\circ }=143^{\circ }$ As a compass bearing this is S37°E (since $180^{\circ }-143^{\circ }=37^{\circ }$.

Always draw a north line at each point and mark the angles you know.

• Labelling opposite and adjacent from the wrong angle.
• Dividing when you should multiply (or the reverse): write the ratio as a fraction first, then rearrange.
• Calculator in radians mode: $\sin 30^{\circ }$ should give 0.5.
• Typing degrees and minutes as a decimal (25°30′ is 25.5°, not 25.3°).
• Putting the angle of depression inside the triangle at the top, where the angle is actually $90^{\circ }$ minus it.
• Measuring true bearings anticlockwise, or not writing three digits (write 065°, not 65°).

• Draw and label a diagram, even if one is given: mark the right angle, the known sides, the angle and the unknown.
• Show the ratio you used, the substitution and the rearrangement. A correct method usually earns marks even after a slip.
• Round only at the end, to the accuracy asked for (often 1 or 2 decimal places, or the nearest degree or minute).
• Check the answer is reasonable: the hypotenuse is the longest side; angles in a right-angled triangle other than the right angle are less than 90°.

• Applications of measurement (Year 11): Pythagoras' theorem is also used there, in perimeter and area problems.
• Time and location (Year 11): directions and positions on the Earth extend the idea of bearings.
• Ratios and rates (Year 12): scale drawings and maps are often combined with bearings and right-angled triangles.