10. Trigonometry and Pythagoras’ Theorem
Trigonometry is a branch of mathematics involving the study of triangles, and has applications in fields such as engineering, surveying, navigation, optics, and electronics. The ability to use and manipulate trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and complex numbers.
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Below is a quiz on trigonometry. If you can answer all these questions, then you can skip this chapter as you already have the requisite knowledge.
Quiz: Trigonometry
1. Find the value of [latex]h[/latex].
2. Find the value of [latex]b[/latex]
3. A fire fighter's ladder is 5m long and must reach 4.8m up a vertical wall. (Assume horizontal ground).
(a) How far from the base of the wall should the foot of the ladder be?
(b) What angle does the ladder make with the ground?
4. For the right-angled triangle below, find the value of [latex]x[/latex].
5. For the triangle below, find the value of [latex]x[/latex].
6. In the triangle below find the value of [latex]x[/latex].
Answers
[latex]\mathbf{1.} \ 10[/latex]
[latex]\mathbf{2.} \ 5[/latex]
[latex]\mathbf{3.} \ a)\ 1.4m\quad b) 73.7^{\circ}[/latex]
[latex]\mathbf{4.} \ 20.78[/latex]
[latex]\mathbf{5.} \ 12.77[/latex]
[latex]\mathbf{6.} \ 11[/latex]
If you need to review this topic continue reading
Right-angled triangles
In a right-angled triangle the three sides are given special names.
The side opposite the right angle is called the hypotenuse (h) – this is always the longest side of the triangle.
The other two sides are named in relation to another known angle (or an unknown angle under consideration).
In fig.10.1 below, the opposite side is the side opposite the given angle [latex]θ[/latex]. The other side is called the adjacent side as it is next to or adjacent to the given angle [latex]θ[/latex].
(Note that the two shorter sides may be labelled either opposite or adjacent, depending on which angle is under consideration.)
Pythagoras' theorem
Pythagoras' Theorem states that for any right-angled triangle the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
This can be written mathematically as [latex]h^2=a^2+b^2[/latex] where [latex]h[/latex] is the length of the hypotenuse and [latex]a[/latex] and [latex]b[/latex] are the lengths of the other two sides of the triangle.
Examples: Pythagoras' theorem
1. Find the length of the hypotenuse in the triangle below.
Solution:
\begin{align*}
h^2&=a^2+b^2 \\
&=6^2+8^2 \\
&=36+64 \\
&=100 \\
\implies h&= \sqrt{100}\\
&=10.
\end{align*}
2. Find the length of the side [latex]a[/latex].
Solution:
\begin{align*}
h^2&=a^2+b^2 \\
\implies 53.85^2&=a^2 + 50^2 \\
\implies 53.85^2-50^2&=a^2 \\
\implies 399.8&=a^2 \\
\implies a&= \sqrt{399.8}\\
&=20.
\end{align*}
Use Pythagoras' Theorem to solve the questions in the exercise below.
Exercises: Pythagoras' theorem
1. Find the value of the unknown side.
(a)
(b)
2. A man walks 5km north and then 5km west. How far away is he from his starting point?
3. A three metre ladder is placed on horizontal ground against a vertical wall, with the foot of the ladder one metre away from the base of the wall. How far up the wall does the ladder reach?
Answers
1. (a) [latex]x=15[/latex] (b) [latex]a=4\sqrt{2}≈5.657[/latex]
2. [latex]5\sqrt{2}≈7.071km[/latex]
3. [latex]2.828m[/latex]
Trigonometric ratios
In a right-angled triangle the following ratios are defined for a given angle [latex]\theta[/latex]
[latex]\text{sine}\,\theta =\dfrac{\text{opposite side length}}{\text{hypotenuse length}}[/latex]
[latex]\text{cosine}\,\theta =\dfrac{\text{adjacent side length}}{\text{hypotenuse length}}[/latex]
[latex]\text{tangent}\,\theta =\dfrac{\text{opposite side length}}{\text{adjacent side length }}[/latex]
These ratios are abbreviated to [latex]\sin\theta , \cos\theta[/latex] , and [latex]\tan\theta[/latex] respectively.
A useful memory aid is SOH CAH TOA:
[latex]\boldsymbol{S}in\,\theta=\dfrac{\boldsymbol{O}pp}{\boldsymbol{H}yp}[/latex] [latex]Cos\,\theta=\dfrac{\boldsymbol{A}dj}{\boldsymbol{H}yp} [/latex] [latex]\boldsymbol{T}an\,\theta=\dfrac{\boldsymbol{O}pp}{\boldsymbol{A}dj}[/latex]
These ratios can be used to find unknown sides and angles in right-angled triangles
Examples: Trigonometric ratios
Evaluating ratios
1. In the right-angled triangle below evaluate:
(a) [latex]\sinθ , \cos\theta[/latex] , and [latex]\tan\theta[/latex]
(b) [latex]\sin\phi , \cos\phi[/latex] , and [latex]\tan\phi[/latex]
Solution:
(a) [latex]\sin\theta=\dfrac{opp}{hyp}[/latex]
[latex]\sinθ=\dfrac{2}{4}=\dfrac{1}{2}=0.5 [/latex]
[latex]\cos\theta=\dfrac{adj}{hyp} [/latex]
[latex]\cosθ=\dfrac{3.464}{4}=0.866[/latex]
[latex]\tan\theta=\dfrac{opp}{adj}[/latex]
[latex]\tan\theta=\dfrac{2}{3.464}=0.577[/latex]
(b) [latex]\sin\phi=\dfrac{opp}{hyp}[/latex]
[latex]\sin\phi=\dfrac{3.464}{4}=0.866 [/latex]
[latex]\cos\phi=\dfrac{adj}{hyp} [/latex]
[latex]\cos\phi=\dfrac{2}{4}=\dfrac{1}{2}=0.5[/latex]
[latex]\tan\phi=\dfrac{opp}{adj}[/latex]
[latex]\tan\phi=\dfrac{3.464}{2}=1.232[/latex]
Finding angles
2. Find the value of the angle (θ) in the triangle below.
Solution
Step 1: Determine which ratio to use.
In this triangle, we know two sides and need to find the angle. The known sides are the opposite side and the hypotenuse. The ratio that relates the opposite side and the hypotenuse is the sine ratio.
\begin{align*}
\sin\theta &=\frac{Opp}{Hyp} \\
&=\frac{13.4}{19.7}\\
& =0.6802\ . \\
\end{align*}
We now use the inverse sine function, denoted [latex]\sin^{-1}[/latex] or [latex]\text{arcsin}[/latex] on your calculator, to "undo" the [latex]\sin[/latex] function and get theta on it's own.
\begin{align*}
\sin ^{-1} \left( \sin \theta \right) &= \sin ^{-1} \left( 0.6802 \right) \\
\implies \theta &= \sin ^{-1} \left( 0.6802 \right) \\
&=42.9^{\circ}.
\end{align*}
Finding side lengths
3. Find the value of the indicated unknown side length in each of the following right-angled triangles.
(a)
Solution
In this problem, we know an angle and the adjacent side.
The side to be determined is the opposite side.
The ratio that relates these two sides is the tangent ratio.
[latex]\tan\theta =\dfrac{Opp}{Adj}[/latex]
[latex]\tan27° ={\dfrac{b}{42}}[/latex]
[latex](\tan27°)\times42 =b[/latex]
[latex]b =21.4cm[/latex]
(b)
Solution
In this problem, we know an angle and the adjacent side.
The side to be determined is the hypotenuse.
The ratio that relates these two sides is the cosine ratio.
[latex]\cos\theta =\dfrac{Adj}{Hyp}[/latex]
[latex]\cos35° ={\dfrac{7}{x}}[/latex]
[latex](\cos35°)\times x =7[/latex]
[latex]x =\dfrac{7}{\cos{35°}}[/latex]
[latex]x =8.55[/latex]
Special angles and exact values
There are some special angles for which the trigonometric functions have exact values rather than decimal approximations.
Applying the rules for sine, cosine and tangent to the triangles below, exact values for the sine, cosine and tangent of the angles [latex]30°, 45°[/latex] and [latex]60°[/latex] can be found.
[latex]\sin45° ={\dfrac{1}{\sqrt{2}}\qquad}\cos\,45°=\dfrac{1}{\sqrt{2}}\qquad \tan\,45°=1[/latex]
[latex]\sin{60°} ={\dfrac{\sqrt{3}}{2}\qquad}\cos\,60°=\dfrac{1}{2}\qquad \tan\,60°=\sqrt{3}[/latex]
[latex]\sin{30°} ={\dfrac{1}{2}\qquad}\cos\,30°=\dfrac{\sqrt{3}}{2}\qquad \tan\,30°=\dfrac{1}{\sqrt{3}}[/latex]
It is worthwhile remembering these values as they frequently arise in practice.
Here are some exercises to try.
Exercises: Trigonometric ratios
1. Using the right-angled triangle below find:
(a) [latex]\sin\theta[/latex]
(b) [latex]\tan\theta[/latex]
(c) [latex]\cos\alpha[/latex]
(d) [latex]\tan\alpha[/latex] (Hint: Use Pythagoras' theorem to find the hypotenuse)
2. Find the value of the indicated unknown (side length or angle) in each of the following diagrams.
a)
b)
c)
d)
3. In a right-angled triangle [latex]\sin\phi = 0.55[/latex] and the hypotenuse is [latex]21[/latex]mm. Find the length of each of the other two sides.
4. A ladder, 4m long, is placed on horizontal ground with the foot of the ladder 1m from the base of a vertical wall.
(a) How far up the wall does the ladder reach?
(b) What angle does the ladder make with the ground?
Answers
1. (a) [latex]\sinθ=\dfrac{12}{13}[/latex] (b) [latex]\tanθ=\dfrac{12}{5}[/latex] (c) [latex]\cosα=\dfrac{12}{13}[/latex] (d) [latex]\tanα= \dfrac{5}{12}[/latex]
2. (a) [latex]x≈8.391cm[/latex] (b) [latex]x≈6.302mm[/latex] (c) [latex]h≈14.197 \,\ β=25°[/latex] (d) [latex]θ=30° \,\ Φ=60° \,\ a= 5\sqrt{3}≈8.66[/latex]
3. side opposite [latex]Φ=11.55mm[/latex] the side adjacent to [latex]Φ=17.54mm[/latex]
4. (a) [latex]\sqrt{15}≈3.873m.[/latex] (b) [latex]75.5°[/latex]
Non-right-angled triangles: Sine and cosine rules
Sine rule
The sine rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:
(i) One side and any two angles
OR
(ii) Two sides and an angle opposite one of the given sides.
In other words, to use the sine rule, we need a triangle where we know at least one "pair" of a side and its opposite angle.
In the triangle [latex]ABC[/latex] below:
angles [latex]∠A,∠B,∠C[/latex], are the angles at the vertices [latex]A,B,C[/latex] respectively,
and [latex]a,b,c[/latex] are the side lengths opposite the angles [latex]∠A,∠B,∠C[/latex] respectively.
The sine rule states:
[latex]\dfrac{a}{\sin{A}} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/latex]
or
[latex]\dfrac{\sin A}{a} =\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/latex]
Note that side [latex]a[/latex] is the side opposite angle [latex]A[/latex], side [latex]b[/latex] is opposite angle [latex]B[/latex] and side [latex]c[/latex] is opposite angle [latex]C[/latex].
Cosine rule
The Cosine Rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:
(i) Two sides and the angle between them
OR
(ii) All three sides of the triangle
The cosine rule is used to find the length of a side opposite a given angle or the angle opposite a given side.
The angles [latex]A, B, C[/latex] are the angles at the vertices [latex]A, B, C[/latex] respectively. The sides [latex]a, b,[/latex] and [latex]c[/latex] are opposite angles [latex]A, B, C[/latex] respectively.
Cosine rule:
[latex]a^{2} =b^{2}+c^{2}-2bc\cos A[/latex]
[latex]b^{2} =a^{2}+c^{2}-2ac\cos B[/latex]
[latex]c^{2} =a^{2}+b^{2}-2ab\cos C[/latex]
N.B. The side on the left hand side of the equation is opposite the angle listed at the end of the equation.
Sine Rule: If you know an angle and the length of the side opposite that angle, you can use the sine rule to find the remaining angles and sides (as long as we know one other side or angle).
Cosine Rule: If you know two sides and the included angle, you can use the cosine rule to find the opposite side (or if we know all three sides we can find the angles).
Examples: Non-right-angled triangles (sine and cosine rules)
1. In triangle [latex]PQR[/latex] find:
a) side length [latex]p[/latex]
b) side length [latex]q[/latex]
Solution:
a) Side length [latex]p[/latex]:
Use the sine rule in the form[latex]\dfrac{p}{\sin P} =\dfrac{r}{\sin R}[/latex]
[latex]\dfrac{p}{\sin70°} =\dfrac{15}{\sin30°}[/latex]
[latex]p =\dfrac{15\times \sin70°}{\sin30°}[/latex]
[latex]p =28.2cm[/latex]
b) Side length [latex]q[/latex]:
Angle [latex]Q[/latex] is found using the fact that the sum of the three interior angles of a triangle add to [latex]180°[/latex].
[latex]∠Q = 180° − (70° + 30°) = 80°[/latex]
From the sine rule [latex]\dfrac{q}{\sin Q} =\dfrac{r}{\sin R}[/latex]
[latex]\dfrac{q}{sin80°} =\dfrac{15}{\sin30°}[/latex]
[latex]q =\dfrac{15\times \\sin80°}{\sin30°}[/latex]
[latex]q =29.6cm[/latex]
2. In triangle [latex]ABC[/latex] find:
a) angle [latex]C[/latex] b) angle [latex]A[/latex] c) side length [latex]a[/latex] (the side opposite angle A).
Solution:
a) Angle [latex]C:[/latex] Use the sine rule in the form [latex]\dfrac{\sin A}{a} =\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/latex]
The relevant part of the formula is [latex]\dfrac{\sin B}{b} =\dfrac{\sin C}{c}[/latex]
[latex]\dfrac{\sin{126°}}{20} =\dfrac{\sin{C}}{12}[/latex]
[latex]\dfrac{\sin126°}{20}\times12 =\sin{C}[/latex]
[latex]\sin{C} =0.485[/latex]
[latex]C =\sin^{-1}0.485[/latex]
[latex]C =29°[/latex]
b) Angle [latex]A[/latex]
[latex]A = 180°-(126°+29°)=25°[/latex]
c) side length [latex]a[/latex]: Use the sine rule in the form [latex]\dfrac{a}{\sin A} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/latex]
The relevant part of the formula is[latex]\dfrac{a}{\sin A} =\dfrac{b}{\sin B}[/latex]
[latex]\dfrac{a}{\sin{25°}} =\dfrac{20}{\sin126°}[/latex]
[latex]a =\dfrac{20}{\sin126°}\times \sin25°[/latex]
[latex]a =10.4[/latex]
3. Given a triangle [latex]ABC[/latex] with angle [latex]A = 30°[/latex], side [latex]c = 12 cm[/latex] and side [latex]a = 8 cm[/latex], find angle [latex]C[/latex].
Solution:
In this case there are two possible solutions, as shown below. This is called the ambiguous case of the sine rule.
Use the sine rule in the form [latex]\dfrac{\sin A}{a} =\dfrac{\sin C}{c}[/latex]
[latex]\dfrac{\sin{30°}}{8} =\dfrac{\sin C}{12}[/latex]
[latex]\dfrac{\sin30°}{8}\times12 =\sin{C}[/latex]
[latex]\sin{C} =0.75[/latex]
[latex]C =\sin^{-1}0.75[/latex]
[latex]C =48.6° or \:131.4°[/latex]
[latex]C = 48.6°[/latex], this solution gives the triangle [latex]ABC'[/latex].
[latex]C = 131.4° = (180° -48.6°)[/latex], this solution gives triangle[latex]ABC[/latex].
Note: In any non-right-angled triangle, where two sides and the non-included angle are given, check for the ambiguous case.
If the angle is acute and the length of the side adjacent to the angle is greater than the length of the side opposite the angle there may be 2 possible solutions.
4. Find the value of [latex]a[/latex] in this triangle:
[latex]a^{2} =b^{2}+c^{2}-2bc\cos A[/latex]
[latex]a^{2} =12^{2}+15^{2}-2\times12\times15\times \cos83°[/latex]
[latex]{a^{2}} =144+225-360{\times}\cos83°[/latex]
[latex]{a^{2}} =369-43.87[/latex]
[latex]a^{2} =325.13[/latex]
[latex]a =18.03[/latex]
5. Find the size of angle [latex]B[/latex] in this triangle:
[latex]b^{2} =a^{2}+c^{2}-2ac\cos B[/latex]
[latex]11^{2} =5^{2}+7^{2}-2\times5\times7\times \cos B[/latex]
[latex]121 =25+49-70\times \cos B[/latex]
[latex]121-25-49 =-70\times \cos B[/latex]
[latex]47 =-70\times \cos B[/latex]
[latex]\dfrac{47}{-70} =\cos B[/latex]
[latex]B =\cos^{-1}\left(-\frac{47}{70}\right)[/latex]
[latex]B =132°11^{\prime}[/latex]
Now try the exercises below.
Exercises: Non-right-angled triangles (sine and cosine rules)
1. For the following triangles, find the unknown sides.
(a)
(b)
2. For the following triangle, find all unknown angles and sides (answer to two decimal places).
3. Given triangle [latex]ABC[/latex], where angle [latex]A = 35°[/latex], side[latex]a = 16[/latex] and side [latex]c = 21,[/latex] find the magnitude of angles[latex]B[/latex] and [latex]C[/latex] and the length of side[latex]b[/latex].
4. Use the sine or cosine rule to find the pronumeral shown:
(a)
(b)
(c)
5. Find the magnitude of the labelled, unknown angle
a)
b)
Answers
1. (a) [latex]b = 18.1, c =19.9[/latex] (b) [latex]q = 53.5, r = 41.2[/latex]
2. [latex]α = 34.34° , β = 25.66°, b = 6.51[/latex]
3. Ambiguous case: [latex]C = 131.2°, B = 13.8°, b = 6.7[/latex] or [latex]C = 48.8°, B = 96.2°, b = 27.7[/latex]
4. a) [latex]28.7[/latex] b) [latex]4.38[/latex] c) [latex]8.33[/latex]
5. a) [latex]36.3⁰[/latex] b) [latex]29.2⁰[/latex]
Key takeaways
1. Pythagoras' Theorem tells us that in a right-angled triangle, the length of the hypotenuse (longest side) squared is equal to the sum of the squares of the other two sides.
[latex]h^2=a^2+b^2[/latex]
2. In a right-angled triangle we can use the sine, cosine and tangent ratios to solve for unknown sides or angles.
If [latex]θ[/latex] is a given angle, then the three sides of the triangle can be labelled as [latex]H[/latex] (hypotenuse), [latex]O[/latex] (opposite side), [latex]A[/latex] (adjacent side)
[latex]\sin\,\theta=\dfrac{O}{H}[/latex] [latex]\cos\,\theta=\dfrac{A}{H} [/latex] [latex]\tan\,\theta=\dfrac{O}{A}[/latex]
A useful memory aid is SOH CAH TOA
3. In any triangle, we can use the sine rule and the cosine rule to solve for unknown sides or angles.
In any triangle [latex]ABC[/latex] the angles [latex]A,B,C[/latex], are the angles at the vertices [latex]A,B,C[/latex] respectively and [latex]a,b,c[/latex] are the side lengths opposite the angles [latex]A,B,C[/latex] respectively, we can apply the the following rules:
Sine Rule:
[latex]\dfrac{a}{\sin{A}} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/latex] or
[latex]\dfrac{\sin A}{a} =\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/latex]
Cosine Rule:
[latex]a^{2} =b^{2}+c^{2}-2bc\cos A[/latex] or
[latex]b^{2} =a^{2}+c^{2}-2ac\cos B[/latex] or
[latex]c^{2} =a^{2}+b^{2}-2ab\cos C[/latex]
