Hello, and welcome to today's mathematics lesson. We are going to explore the solution of right-angled triangles. By the end of this lesson, you will understand how to find unknown sides and angles in right-angled triangles using trigonometry, and you will see how these skills apply to real-world problems involving heights, distances, and geometric shapes.
Let us begin with the fundamental idea. To solve a right-angled triangle means to find all the remaining sides and angles when some information is already given. There are two standard situations you will encounter.
First, when one side and one acute angle are given. Second, when two sides of the triangle are given. Once you know two elements, you can find everything else using trigonometric ratios.
Recall the three basic ratios. In a right-angled triangle, if you mark the angle you are working with, then the side opposite to it is called the perpendicular, the side next to it is the base, and the longest side opposite the right angle is the hypotenuse.
Sine of an angle equals perpendicular divided by hypotenuse. Cosine equals base divided by hypotenuse. Tangent equals perpendicular divided by base. And remember, cotangent is simply base divided by perpendicular.
Here is the golden rule that will guide all your calculations. When one side and one acute angle are known, set up a ratio where the unknown side divided by the known side equals the appropriate trigonometric ratio of the given angle.
For example, suppose in triangle ABC, right-angled at B, side BC equals 20 centimetres and angle A equals 30 degrees. To find AB, notice that AB is adjacent to angle A, while BC is opposite to it. Therefore, AB over BC equals cotangent of angle 30 degrees. This gives AB over 20 equals √3, so AB equals 20 √3, which is approximately 34.64 centimetres.
Sometimes you need to find angles rather than sides. When you know two sides, form the appropriate ratio and identify which angle has that trigonometric value.
Imagine a right triangle where the perpendicular is 5 units and the hypotenuse is 10 units. Sine of angle x equals 5 over 10, which is one-half. Since sine of angle 30 degrees equals one-half, angle x must be 30 degrees.
Similarly, if the perpendicular and base are both 10 units, then tangent of angle y equals 10 over 10, which is 1. Since tangent of angle 45 degrees equals 1, angle y equals 45 degrees.
Many problems involve combining two right triangles that share a common side. This shared side becomes the bridge that connects the two triangles.
Consider a figure where point D lies on line BC, and from A, perpendicular AD is drawn. If angle ABD is 45 degrees and AD equals 10 centimetres, then in triangle ABD, tangent of angle 45 degrees equals AD over BD. Since tangent of 45 degrees is 1, BD equals 10 centimetres.
Now look at triangle ADC where AC equals 20 centimetres. Sine of angle C equals AD over AC, which is 10 over 20, or one-half. Therefore angle C equals 30 degrees. Using tangent of angle 30 degrees equals AD over DC, you can find DC equals 10 √3, approximately 17.32 centimetres. Finally, BC equals BD plus DC, giving 27.32 centimetres.
Some problems require setting up equations with unknowns. Suppose you need to find height AC where C is vertically below B, and D is a point on the ground such that BC equals 40 metres plus DC.
From two different observation points, you have tangent of angle 30 degrees equals AC over BC, and tangent of angle 45 degrees equals AC over DC. From the second equation, DC equals AC. Substituting into the first equation gives 40 plus AC equals AC times √3. Rearranging, AC times (√3 - 1) equals 40. Therefore AC equals 40 divided by (√3 - 1).
Rationalising the denominator by multiplying numerator and denominator by (√3 + 1), you get AC equals 40 times (√3 + 1) divided by 2, which simplifies to 20 times 2.732, giving approximately 54.64 metres.
Let us apply these skills to properties of special quadrilaterals. In a rhombus, the diagonals bisect each other at right angles and also bisect the vertex angles.
Consider rhombus ABCD with each side 10 centimetres and angle A equal to 60 degrees. The diagonals intersect at O, creating four congruent right triangles. Angle OAB equals 30 degrees, half of angle A.
In triangle AOB, sine of angle 30 degrees equals OB over AB, so OB equals 5 centimetres. Cosine of angle 30 degrees equals OA over AB, so OA equals 5 √3, approximately 8.66 centimetres. Therefore diagonal BD equals 10 centimetres, and diagonal AC equals 17.32 centimetres.
Real-world applications often involve angles of elevation and paths that change direction.
Imagine a rocket launched from point P. It rises 40 kilometres vertically to point A, then travels another 40 kilometres at 60 degrees to the vertical to reach point B. Point C is directly below B at ground level.
Draw AD perpendicular to BC. The angle BAD equals 30 degrees, since 90 minus 60 equals 30. In triangle ABD, sine of angle 30 degrees equals BD over 40, so BD equals 20 kilometres. The total height BC equals 40 plus 20, which is 60 kilometres.
For the horizontal distance, cosine of angle 30 degrees equals AD over 40, so AD equals 20 √3, approximately 34.64 kilometres. Since PC equals AD, the horizontal distance is 34.64 kilometres.
Problems with multiple perpendicular lines require careful identification of which triangle contains which elements.
Suppose AB is perpendicular to BC, DC is perpendicular to BC, and BD is perpendicular to AC. Given angle D equals 30 degrees and DC equals 60 √3 metres, you need to find AB.
First, in triangle BCD, tangent of angle 30 degrees equals BC over DC. This gives BC equals 60 metres. Angle DBC equals 60 degrees, found from the angle sum of triangle BCD.
Now, angle BOC equals 90 degrees, so angle BCO equals 30 degrees. Finally, in triangle ABC, tangent of angle 30 degrees equals AB over BC. Thus AB equals 60 over √3, which rationalises to 20 √3 metres, or approximately 34.64 metres.
Trapezium problems combine parallel lines with right triangle constructions.
Consider trapezium ABCD with AB parallel to DC, angle C equals 120 degrees, DC equals 28 centimetres, and BC equals 40 centimetres. Since AB and DC are parallel, angle B equals 60 degrees, because co-interior angles sum to 180 degrees.
Draw CE perpendicular to AB. In triangle CBE, sine of angle 60 degrees equals CE over 40, so CE equals 20 √3 centimetres. Cosine of 60 degrees equals BE over 40, so BE equals 20 centimetres.
Therefore AB equals BE plus AE, which equals 20 plus 28, giving 48 centimetres. AD equals CE, which is 20 √3 centimetres. The area equals half times (AB + DC) times height, which equals half times 76 times 20 √3, giving 760 √3 square centimetres, or approximately 1316.32 square centimetres.
Let us recap the key takeaways from this lesson.
First, to solve a right-angled triangle, you need either one side and one acute angle, or two sides.
Second, always set up the ratio of unknown side to known side equal to the appropriate trigonometric ratio of the given angle.
Third, when two right triangles share a common side, use that side as a bridge to connect information from both triangles.
Fourth, in a rhombus, the diagonals bisect each other at right angles and bisect the vertex angles, creating useful right triangles.
Fifth, for angles of elevation and real-world paths, carefully track which angle is measured from which reference line.
Sixth, always rationalise denominators containing surds to obtain numerical answers.
Remember, trigonometry is a powerful tool that connects angles with lengths. With practice, you will recognise patterns quickly and choose the most efficient ratio for each situation. Keep your diagrams neat, label all known quantities, and work step by step. You have got this. Until next time, keep exploring and enjoy your mathematics.