11.08 Right angled and non-right angled triangles

In Chapter 7, the  area of a triangle  was calculated using a simple formula, providing the base length of the triangle is known as well as the perpendicular height. \text{Area}= \dfrac{1}{2}\,bh is the formula, which uses the base, b, and height, h.

Sometimes the height is referred to as the altitude of the triangle and it must always be perpendicular to the base.

However, consider the triangle given below, where no perpendicular height measurement is known.

If two sides lengths and their included interior angle (SAS) is known, there is another formula that can be used to find the area, which uses the sine ratio. The formula is most commonly written as follows.

For a triangle with side lengths a and b and an included angle C, then the area of the triangle is given by: \text{Area}= \dfrac{1}{2}\, ab\,\sin C

Examples

Example 1

Calculate the area of the following triangle. Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the sine area rule: \text{Area}=\dfrac{1}{2} \, ab\, \sin C

Apply the idea

\displaystyle \text{Area}	\displaystyle =	\displaystyle \dfrac{1}{2} \, ab\, \sin C	Use sine area rule
	\displaystyle =	\displaystyle \dfrac{1}{2} \times 6 \times 2.2 \times \sin 73\degree	Substitute values
	\displaystyle =	\displaystyle 6.31 \text{ m}^2	Evaluate

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In this chapter, it was discovered that for right-angled triangles, where one known angle is 90\degree, the following rules can be used:

For right-angled triangles

• Sides lengths must always satisfy  Pythagoras' theorem  , c^2 = a^2 + b^2

•  Trigonometric ratios  of tangent, sine and cosine exist, \\ \tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}, \, \cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}, \, \sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}

• Area of the triangle can be found with simple formula A = \dfrac{1}{2}\,bh, using the triangle's base length b and perpendicular height h (i.e. the two shorter sides).

In this chapter, it was discovered that for non right-angled triangles (acute, obtuse), when considering triangles with no 90\degree angle, the following rules can be used:

For all other triangles

•  Cosine rule  , c^2 = a^2 + b^2 - 2ab \cos C, is valid for acute, right-angled & obtuse triangles

•  Sine rule  , \dfrac{a}{\sin A} = \dfrac{b}{\cos B} = \dfrac{c}{\tan C} is also valid for acute, right-angled & obtuse triangles (including the ambiguous case)

Area of any triangle with a base length b and perpendicular height h can be calculated using the simple formula, A = \dfrac{1}{2}\,bh.

• Area of any triangle with two known side lengths a & b and an included angle C (side-angle-side) can be calculated using the formula, A = \dfrac{1}{2}ab \sin C.

• Area of any triangle with all three known side lengths (side-side-side) a,\, b, & c can be calculated using Heron's formula A = \sqrt{s(s-a)(s-b)(s-c)} where s is the semi-perimeter of the triangle, s = \dfrac{a+b+c}{2}.

So when given a problem to solve that relates to triangles, the following steps can be followed:

1. Identify whether the triangle is acute, right-angled or obtuse.

2. If the triangle is right-angled, the first set of rules above can be used to determine missing side lengths and/or angles. The area of the triangle can also be calculated using the simple formula given above.

3. If the triangle has no right angles, the second set of rules above can be used to determine missing side lengths and/or angles. The area of the triangle can also be calculated using one of the three formulae given above, depending on what information about the triangle is given.

Examples

Example 2

Find the side length a using the sine rule. Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}.

Apply the idea

\displaystyle \frac{a}{\sin 33\degree}	\displaystyle =	\displaystyle \frac{18}{\sin 69\degree }	Substitute b=18, \, A=33, \, B=69
\displaystyle a	\displaystyle =	\displaystyle \frac{18}{\sin 69\degree} \times \sin 33\degree	Multiply both sides by \sin 33\degree
	\displaystyle =	\displaystyle 10.50	Evaluate

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Example 3

A boat travels \text{S}\, 14 \degree \text{E} for 12 \text{ km} and then changes direction to \text{S}\, 49 \degree \text{E} for another 16 \text{ km}.

a

Find x, the distance of the boat from its starting point. Give your answer to two decimal places.

Worked Solution

Create a strategy

Find the opposite angle and then use the cosine rule to find x.

Apply the idea

The angle opposite x, let's call it \theta, is made up of two angles. The smaller angle is alternate to the angle of 14\degree since the two arrows pointing north are parallel. So the small angle equals 14\degree. The larger angle is on a straight line with 49\degree , so they add up to 180\degree.

\displaystyle \text{Larger angle}	\displaystyle =	\displaystyle 180-49	(Angles on a straight line)
	\displaystyle =	\displaystyle 131\degree	Evaluate
\displaystyle \theta	\displaystyle =	\displaystyle 131+14	Add the two angles
	\displaystyle =	\displaystyle 145\degree	Evaluate

Now we can use the cosine rule to find x.

\displaystyle c^2	\displaystyle =	\displaystyle a^2+b^2-2ab \cos C	Use the cosine rule
\displaystyle x^2	\displaystyle =	\displaystyle 12^2+16^2 -2 \times 12 \times 16 \times \cos 145\degree	Substitute c=x,a=12,b=16, C =145\degree
	\displaystyle =	\displaystyle 714.554\,385	Evaluate
\displaystyle x	\displaystyle =	\displaystyle \sqrt{714.554\,385}	Take the square root of both sides
	\displaystyle \approx	\displaystyle 26.73 \text{ km}	Evaluate and round

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b

Find the angle b as labelled in the diagram. Express your answer to the nearest degree.

Worked Solution

Create a strategy

Use the sine rule.

Apply the idea

\displaystyle \dfrac{\sin A}{a}	\displaystyle =	\displaystyle \dfrac{\sin C}{c}	Use the sine rule
\displaystyle \dfrac{\sin b}{16}	\displaystyle =	\displaystyle \dfrac{\sin 145}{26.73}	Substitute A=b, a=16, C=145, c=26.73
\displaystyle \sin b	\displaystyle =	\displaystyle \dfrac{\sin 145}{26.73} \times 16	Multiply both sides by 16
	\displaystyle \approx	\displaystyle 0.34333	Evaluate
\displaystyle b	\displaystyle =	\displaystyle \sin ^{-1}(0.34333)	Take the inverse sine of both sides
	\displaystyle \approx	\displaystyle 20 \degree	Evaluate and round

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c

Hence write down the bearing that the boat should travel on to return to the starting point.

Worked Solution

Create a strategy

Find the angle, a, by using the alternate angles on parallel lines.

Apply the idea

Angle a is alternate to the angle (b+14)\degree at the starting point, because both north lines are parallel. So these angles are equal.

\displaystyle a	\displaystyle =	\displaystyle b+14	Add the two angles
	\displaystyle =	\displaystyle 20+14	Substitute b=20
	\displaystyle =	\displaystyle 34\degree	Evaluate

So the compass bearing is \text{N} \, 34\degree \text{W}.

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