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Summary

Solving triangle problems

In a nutshell

Triangle problems can involve using a combination of different rules concerning area, trigonometry and Pythagoras' theorem. You will need to know how to combine different equations and rules in order to find missing sides and angles in triangles.

Equations

Rule	equations
Trigonometric ratios	$\sin(\theta) = \dfrac{opposite}{hypotenuse}$
	$\cos(\theta)= \dfrac{adjacent}{hypotenuse}$
	$\tan(\theta) = \dfrac{opposite}{adjacent}$
Area of a triangle	$Area = \dfrac12 bh$
Area of a triangle	$Area = \dfrac12 ab\sin(C)$
The cosine rule	$a^2=b^2+c^2-2bc\cos(A)$
The cosine rule	$\cos(A) = \dfrac{b^2+c^2-a^2}{2bc}$
The sine rule	$\dfrac{a}{\sin(A)}=\dfrac{b}{\sin(B)}=\dfrac{c}{\sin(C)}$
The sine rule	$\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}$
Pythagoras' theorem	$a^2+b^2=c^2$

Triangle problems

There are a variety of triangle rules and equations you will need to use to find a missing angle or side. You will also need to be able to combine these equations algebraically to calculate unknown values.

Example 1

In the triangle below, find the values of $x$ , $y$ and $z$ to one decimal place.

Maths; Trigonometry; KS5 Year 12; Solving triangle problems

$\triangle ABC$ has three sides given which can be used to find a missing angle using the cosine rule:

$\cos(A) = \dfrac{b^2+c^2-a^2}{2bc}$

Find $z$ :

$\begin{aligned} \cos(z) &= \dfrac{(14.8)^2+(19.6)^2-(11.3)^2}{2(14.8)(19.6)} \\ \\ z&=\cos^{-1}(\dfrac{(14.8)^2+(19.6)^2-(11.3)^2}{2(14.8)(19.6)}) \\ \\ z &= 34.953...^{\circ} \end{aligned}$

$\underline{z = 35.0^{\circ} \ (1 \ d.p.)}$

Find $x$ :

$\begin{aligned} \cos(x) &= \dfrac{(11.3)^2+(19.6)^2-(14.8)^2}{2(11.3)(19.6)} \\ \\ x&=\cos^{-1}(\dfrac{(11.3)^2+(19.6)^2-(14.8)^2}{2(11.3)(19.6)}) \\ \\ x &= 48.622...^{\circ} \end{aligned}$

$\underline{x = 48.6^{\circ} \ (1 \ d.p.)}$

Note: Remember to adjust the variables to fit the angle you are trying to calculate.

Use trigonometric ratios inside the right-angled triangle $\triangle ADC$ to solve for $y$ :

$\sin(\theta) = \dfrac{opposite}{hypotenuse}$

Note that $y$ is the hypotenuse, and the opposite length and the angle have been given and found respectively:

$\begin{aligned} \sin(34.953...) &=\dfrac{y}{14.8} \\ \sin(34.953...) \times 14.8&={y} \\ 8.479...cm&=y \end{aligned}$ 

$\underline{y = 8.5 cm \ (1\ d.p.)}$

Example 2

In the triangle below, given that $ADC$ is a straight line, find $m$ to one decimal place.

$\triangle BDC$ has two sides which are the same length and an angle which is not $60^{\circ}$ . So, it must be an isosceles triangle:

$\angle DBC = \angle BCD= 55^{\circ}$

Angles in a triangle add up to $180^{\circ}$ :

$\angle BDC =180-55-55 = 70\degree$

$ADC$ is a straight line. Angles on a straight line add up to $180^{\circ}$ :

$\angle ADB = 180 - 70 = 110^{\circ}$

In $\triangle ADB$ , two sides have been given: $BA$ and $BD$ . An angle has also been found. This means that the cosine rule can be used to find the missing side:

$a^2=b^2+c^2-2bc\cos(A)$

Substitute values into the cosine rule where $\angle ADB$ is the focus:

$(8)^2 = (5)^2 + m^2 - 2(5)m\cos(110)$ 

$0 = x^2 -2(5)m\cos(110) +(5)^2-(8)^2$

$m^2 - 10\cos(110)(m) -39 = 0$

Solve the quadratic to find $m$ :

$m= \dfrac{-(10\cos(110)) \pm \sqrt{(- 10\cos(110))^2-4(1)(-39)}}{2(1)}$

$\begin{aligned} m&= -4.764... \\ m&=8.185... \end{aligned}$

$m$ is a side length so it must be a positive value:

$\underline{m= 8.2 cm \ (1\ d.p.)}$

Learn with Basics

Length:

Unit 1

Finding missing lengths in a triangle

Unit 2

Sine and cosine rules - Higher

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Optional

Unit 3