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E6.2 Right-angled triangles

Syllabus
0580–2028–2029
Topic
E6.2
Level
Extended

Learning objectives

E6.2.1Know and use the sine, cosine and tangent ratios for acute angles in• Know and use the sine, cosine and tangent ratios for acute angles in calculations involving sides and angles of a right-angled triangle.E6.2.2Solve problems in two dimensions using Pythagoras’ theorem and• Solve problems in two dimensions using Pythagoras’ theorem and trigonometry.E6.2.3Know that the perpendicular distance from a point to a line is the• Know that the perpendicular distance from a point to a line is the shortest distance to the line.E6.2.4Carry out calculations involving angles of elevation and depression• Carry out calculations involving angles of elevation and depression. Angles will be given in degrees and answers should be written in degrees, with decimals correct to one decimal place. Knowledge of bearings may be required.

Choose sine, cosine or tangent

Sine, cosine and tangent compare side lengths relative to one acute angle in a right-angled triangle. The hypotenuse is opposite the right angle; opposite and adjacent depend on the chosen angle.

Ratio Relationship Relevant sides
sine sin⁡θ=OH\sin\theta=\dfrac{O}{H} opposite and hypotenuse
cosine cos⁡θ=AH\cos\theta=\dfrac{A}{H} adjacent and hypotenuse
tangent tan⁡θ=OA\tan\theta=\dfrac{O}{A} opposite and adjacent

Mark the right angle, circle the reference angle and label O, A and H. Select the ratio containing the known and unknown sides. Rearrange it to find a side; use the matching inverse function to find an angle. Keep the calculator in degree mode and retain unrounded values until the final answer.

If H=21.8H=21.8 cm and θ=56∘\theta=56^\circ, then O=21.8sin⁡56∘=18.1O=21.8\sin56^\circ=18.1 cm. If instead O=11O=11 cm and A=27A=27 cm, θ=tan⁡−1(11/27)=22.2∘\theta=\tan^{-1}(11/27)=22.2^\circ to one decimal place.

Adjacent is the non-hypotenuse side beside the chosen angle, so O and A can swap when the reference angle changes. Use inverse trig only for an unknown angle. These three ratios apply here to acute angles in right-angled triangles.

Sequence two-dimensional right-triangle problems

A two-dimensional problem may contain more than one right triangle. Treat each triangle as one step: a result from the first becomes a known value in the next.

Information in the current right triangle Method
two side lengths Pythagoras' theorem
one acute angle and one side sine, cosine or tangent
two sides and an angle required inverse sine, cosine or tangent

Sketch and label the diagram, mark the right angles, and identify which unknown must be found first. Work through connected right triangles in dependency order. Keep full calculator values between stages; rounding an intermediate side can move the final result.

Suppose a first right triangle has rise 66 m and horizontal run 88 m, so its diagonal is 62+82=10\sqrt{6^2+8^2}=10 m. If that diagonal is the adjacent side of a second right triangle with angle 35∘35^\circ, the new opposite side is 10tan⁡35∘=7.0010\tan35^\circ=7.00 m. Each ratio uses sides from one triangle only.

Do not combine lengths from different triangles in one formula. Pythagoras requires a right angle, and the basic sine, cosine and tangent ratios require a right triangle. Non-right triangles need the later sine or cosine rule.

Find the shortest distance from a point to a line

The shortest distance from a point to a line is the length of the perpendicular segment from the point to that line. Its foot meets the line at 90∘90^\circ.

Draw or identify the perpendicular from the point to the line and mark its foot. This creates a right triangle. Choose sine, cosine, tangent or Pythagoras using only that triangle, then report the perpendicular length—not a sloping side or a distance to an endpoint.

\text{Area of triangle}=\frac12\times\text{base}\times\text{perpendicular height}

If a segment of length 8080 m meets the target line at an angle of 72∘72^\circ, the perpendicular distance is opposite that angle: d=80sin⁡72∘=76.1d=80\sin72^\circ=76.1 m. The same distance could be found from area when the triangle's area and base are known.

Shortest means perpendicular to the infinite line. The perpendicular foot may lie on an extension beyond a drawn segment. A sloping connector can never be shorter than the perpendicular distance.

Calculate angles of elevation and depression

An angle of elevation is measured upward from an observer's horizontal line to the line of sight. An angle of depression is measured downward from that horizontal line.

Draw a horizontal through the observer and a vertical height difference at the object. These form a right triangle with the line of sight. Use the height difference—not automatically the object's full height—and the horizontal separation in the appropriate trig ratio. Parallel horizontal lines make an angle of depression equal to the corresponding angle of elevation.

Two vertical poles are 3333 m apart. Their relevant points are 12.612.6 m and 1.51.5 m above level ground, so the vertical difference is 12.6−1.5=11.112.6-1.5=11.1 m. The angle of elevation is tan⁡−1(11.1/33)=18.6∘\tan^{-1}(11.1/33)=18.6^\circ to one decimal place.

If a bearing is supplied, measure it clockwise from north using three figures and use angle facts to locate the horizontal direction before forming the right triangle. Keep the calculator in degree mode and round a decimal angle to one decimal place unless instructed otherwise.

Elevation and depression are measured from a horizontal, not from a vertical pole or the line of sight itself. Subtract observer height when the observer is above ground, and keep intermediate lengths unrounded.