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C6.2 Right-angled triangles

Syllabus
0580–2028–2029
Topic
C6.2
Level
Core

Learning objectives

C6.2.1Know 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.C6.2.2Solve problems in two dimensions using Pythagoras’ theorem and• Solve problems in two dimensions using Pythagoras’ theorem and trigonometry. 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.

Use sine, cosine and tangent in right triangles

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 Use when the known and unknown involve
sine sin⁡heta=extoppositeexthypotenuse\sin heta=\dfrac{ ext{opposite}}{ ext{hypotenuse}} opposite and hypotenuse
cosine cos⁡heta=extadjacentexthypotenuse\cos heta=\dfrac{ ext{adjacent}}{ ext{hypotenuse}} adjacent and hypotenuse
tangent anheta=extoppositeextadjacentan heta=\dfrac{ ext{opposite}}{ ext{adjacent}} opposite and adjacent

Mark the right angle, circle the reference angle, then label O, A and H. Choose the ratio containing the known side and the unknown. To find a side, rearrange before evaluating. To find an angle, use the matching inverse function, such as heta=an−1(O/A)heta= an^{-1}(O/A). Keep the calculator in degree mode and round an angle to one decimal place unless told otherwise.

In a right triangle with hypotenuse 1515 cm and angle 38∘38^\circ, the opposite side is 15sin⁡38∘=9.23…15\sin38^\circ=9.23\ldots cm. If instead the opposite and adjacent sides are 55 cm and 88 cm, heta=an−1(5/8)=32.0∘heta= an^{-1}(5/8)=32.0^\circ to one decimal place.

Adjacent means the non-hypotenuse side beside the chosen angle; it changes when the reference angle changes. Use inverse trig only when the angle is unknown. These ratios apply here to acute angles in right-angled triangles.

Solve two-dimensional right-triangle problems

A two-dimensional trigonometry problem may hide several right triangles. Solve them in an order that turns each new length or angle into data for the next triangle.

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 acute angle required inverse sine, cosine or tangent

Sketch and label the geometry, then mark every right angle. For bearings, draw parallel north lines and measure the three-figure bearing clockwise from north; use angle facts to obtain the triangle's interior angle. Solve one right triangle at a time, carrying unrounded calculator values into later steps. State angles in degrees and give decimal angles to one decimal place.

A first right triangle has horizontal run 1818 m and angle 35∘35^\circ, so its rise is 18an35∘=12.603…18 an35^\circ=12.603\ldots m. That rise is perpendicular to a 2424 m length in a second right triangle, giving a diagonal 242+12.603…2=27.1\sqrt{24^2+12.603\ldots^2}=27.1 m. Using the unrounded rise avoids drift.

Do not apply one ratio across sides that belong to different triangles. A bearing is referenced to north, not automatically to a horizontal edge. Keep intermediate values unrounded; round only the requested final answer. Non-right sine and cosine rules belong to Extended content.