Calculating the area of a non right-angled triangle
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. $A=\frac{1}{2}bh$A=12bh is the formula, which uses the base, $b$b, and height, $h$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 $a$a and $b$b and an included angle $C$C, then the area of the triangle is given by:
$Area=\frac{1}{2}ab\sin C$Area=12absinC
Practice questions
Question 1
Calculate the area of the following triangle.
Round your answer to two decimal places.
Which rule for which triangle?
In this chapter, it was discovered that for right-angled triangles, where one known angle is $90^\circ$90°, the following rules can be used:
- Sides lengths must always satisfy Pythagoras' theorem, $c^2=a^2+b^2$c2=a2+b2
- Trigonometric ratios of tangent, sine and cosine exist $\tan\theta=\frac{opposite}{adjacent},\cos\theta=\frac{adjacent}{hypotenuse},\sin\theta=\frac{opposite}{hypotenuse}$tanθ=oppositeadjacent,cosθ=adjacenthypotenuse,sinθ=oppositehypotenuse
- Area of the triangle can be found with simple formula $A=\frac{1}{2}bh$A=12bh, using the triangle's base length $b$b and perpendicular height $h$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^\circ$90° angle, the following rules can be used:
- Cosine rule, $c^2=a^2+b^2-2ab\cos C$c2=a2+b2−2abcosC, is valid for acute, right-angled & obtuse triangles
- Sine rule, $\frac{a}{\sin A}=\frac{b}{\cos B}=\frac{c}{\tan C}$asinA=bcosB=ctanC is also valid for acute, right-angled & obtuse triangles (including the ambiguous case)
- Area of any triangle with a base length $b$b and perpendicular height $h$h can be calculated using the simple formula $A=\frac{1}{2}bh$A=12bh
- Area of any triangle with two known side lengths $a$a & $b$b and an included angle $C$C (side-angle-side) can be calculated using the formula $A=\frac{1}{2}ab\sin C$A=12absinC
- Area of any triangle with all three known side lengths (side-side-side) $a$a, $b$b & $c$c can be calculated using Heron's formula $A=\sqrt{s(s-a)(s-b)(s-c)}$A=√s(s−a)(s−b)(s−c) where $s$s is the semi-perimenter of the triangle $s=\frac{a+b+c}{2}$s=a+b+c2
So when given a problem to solve that relates to triangles, the following steps can be followed
- Identify whether the triangle is acute, right-angled or obtuse.
- 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.
- 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.
Practice questions
Question 2
Solve for the length of side $a$a.
Round your answer to two decimal places.
Question 3
A boat travels $S$S$14^\circ$14°$E$E for $12$12 km and then changes direction to $S$S$49^\circ$49°$E$E for another $16$16 km.
Find $x$x, the distance of the boat from its starting point. Give your answer to two decimal places.
Find the angle $b$b as labelled in the diagram. Express your answer to the nearest degree.
Hence write down the bearing that the boat should travel on to return to the starting point.
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