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Triangle geometry

One triangle. Three rules.

SOHCAHTOA only works with a right angle. Most triangles haven't got one. So take an ordinary triangle — sides 88 and 55 with 60∘60^\circ between them — and watch where the general rules come from.

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The labelling

Capitals for angles, lower case for sides — and each side takes the name of the angle opposite it. Side aa sits across from angle AA. Every rule in this chapter pairs them exactly that way, so get the labels right and the formulas follow.

The sine rule

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

One altitude, measured two ways, and the rule falls out. Use it when you know a side and the angle opposite it — each side paired with its own angle.

The cosine rule

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A

At A=90∘A = 90^\circ, cos⁡A=0\cos A = 0 and this is Pythagoras. The −2bccos⁡A-2bc\cos A is the correction for every other angle: it shrinks aa when AA is acute and stretches it when AA is obtuse. Back at 60∘60^\circ, a2=64+25−40=49a^2 = 64 + 25 - 40 = 49, so a=7a = 7.

Area

Area=12absin⁡C\text{Area} = \tfrac{1}{2}ab\sin C

Two sides and the angle between them. The altitude is back: its height is 8sin⁡60∘8\sin 60^\circ, so the area is 12×5×8sin⁡60∘=103≈17.3\tfrac{1}{2} \times 5 \times 8\sin 60^\circ = 10\sqrt{3} \approx 17.3. Same idea as half base × height — the sine just finds the height for you.

Two answers

Given A=40∘A = 40^\circ, b=12b = 12 and a=9a = 9: swing a side of length 9 from CC and it meets the base line twice. sin⁡B=0.857\sin B = 0.857 gives B=59.0∘B = 59.0^\circ — but sin⁡(180∘−B)\sin(180^\circ - B) is the same, so B=121.0∘B = 121.0^\circ fits too. Always check for the second triangle.

Chapter 8

Sides meet their opposite angles.

Finding angles as well as sides, bearings problems, and twelve graded exam questions with full worked solutions are waiting.