Triangles without a right angle
Tonight's idea
SOHCAHTOA needs a right angle. Most triangles do not have one, so make one. Label the triangle so that side \(a\) is opposite angle \(A\), and so on. Drop a height \(h\) from \(A\) to the base.
In the left triangle, \(h = b\sin C\) and \(CD = b\cos C\). That leaves \(DB = a - b\cos C\). Now use Pythagoras in the right triangle:
\[c^2 = (b\sin C)^2 + (a - b\cos C)^2 = a^2 + b^2(\sin^2 C + \cos^2 C) - 2ab\cos C.\]
Since \(\sin^2 C + \cos^2 C = 1\):
Cosine rule. \(c^2 = a^2 + b^2 - 2ab\cos C\). Pythagoras, with a correction for how far \(C\) is from a right angle.
The same height read from the other side is \(h = c\sin B\). So \(b\sin C = c\sin B\), and nothing is special about this pair of sides:
Sine rule. \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\).
Both rules hold for obtuse triangles too, by the same steps. Use the cosine rule given two sides and the angle between them, or three sides; the sine rule given a side and its opposite angle.
Watch one: two sides and the angle between
\(a = 7\), \(b = 5\), \(C = 60^\circ\). No side is paired with its angle yet, so start with the cosine rule:
\[c^2 = 49 + 25 - 2(7)(5)\cos 60^\circ = 74 - 35 = 39, \qquad c = \sqrt{39} \approx 6.24.\]
Now \(c\) and \(C\) are a pair, so the sine rule gives \(B\):
\[\sin B = \frac{b\sin C}{c} = \frac{5\sin 60^\circ}{\sqrt{39}} \approx 0.693, \qquad B \approx 43.9^\circ.\]
Angles add to 180°, so \(A \approx 76.1^\circ\). The longest side faces the largest angle, as it should.
Type in three known parts and read the working: mathdetours.com/demos/trianglesolver.html.
Your turn
- In a triangle, \(A = 40^\circ\), \(B = 75^\circ\) and \(a = 10\). Find \(b\).
- A triangle has sides 5, 6 and 7. Find the angle opposite the side of length 7.
- Two boats leave a harbour at the same time on straight courses \(50^\circ\) apart. After an hour one has gone 8 km and the other 11 km. How far apart are they?
- What does the cosine rule say when \(C = 90^\circ\)? When \(C\) is bigger than 90°, is \(c^2\) more or less than \(a^2 + b^2\), and why? Without finding any angle, decide whether the triangle with sides 4, 5 and 7 has an obtuse angle.
Stretch: the parallelogram law
A parallelogram has sides \(a\) and \(b\). Neighbouring angles add to 180°, and \(\cos(180^\circ - C) = -\cos C\). Use the cosine rule on each diagonal to show that the squares of the two diagonals add up to \(2a^2 + 2b^2\).
Answers
- \(\dfrac{b}{\sin 75^\circ} = \dfrac{10}{\sin 40^\circ}\), so \(b = \dfrac{10\sin 75^\circ}{\sin 40^\circ} \approx 15.0\).
- \(49 = 25 + 36 - 2(5)(6)\cos C\), so \(\cos C = \tfrac{12}{60} = 0.2\) and \(C \approx 78.5^\circ\).
- Two sides and the angle between: \(d^2 = 8^2 + 11^2 - 2(8)(11)\cos 50^\circ \approx 185 - 113.1 = 71.9\), so \(d \approx 8.5\) km.
- \(\cos 90^\circ = 0\), so the correction vanishes and it is Pythagoras: \(c^2 = a^2 + b^2\). Past 90° the cosine is negative, so \(-2ab\cos C\) is positive and \(c^2\) is more than \(a^2 + b^2\): opening the angle pushes the far ends apart. For 4, 5, 7: \(4^2 + 5^2 = 41\), which is less than \(7^2 = 49\), so the angle opposite the 7 is obtuse (about 101.5°).
Stretch. Each diagonal is the third side of a triangle with sides \(a\) and \(b\). One faces the angle \(C\), the other faces \(180^\circ - C\). So \(p^2 = a^2 + b^2 - 2ab\cos C\) and \(q^2 = a^2 + b^2 + 2ab\cos C\). Add: the correction terms cancel and \(p^2 + q^2 = 2a^2 + 2b^2\). For a rectangle the diagonals are equal and this is Pythagoras again.