The sine and cosine rules

When the triangle isn't right-angled

• Pythagoras and SOH-CAH-TOA only work for right-angled triangles.
• For any other triangle, you need the sine rule 正弦定理 or the cosine rule 余弦定理.

The sine rule

• For any triangle with sides $a, b, c$ opposite angles $A, B, C$:

• Use it when you know: a side and its opposite angle, plus one other piece.

$y=\sin x$ and $y=\cos x$ are smooth waves between $-1$ and $1$; cosine is sine shifted left by $90^\circ$

The cosine rule

• Use it when you know: two sides and the included angle 夹角 (to find the third side), or all three sides (to find an angle).

In any triangle, sides $a,b,c$ lie opposite angles $A,B,C$; the sine and cosine rules use these

Area of any triangle

• This works for any triangle (not just right-angled ones), using two sides and the included angle.

Worked example

• $b = 7$, $c = 8$, included angle $A = 40^{\circ}$.
• $a^2 = 7^2 + 8^2 - 2(7)(8)\cos 40^{\circ} = 49 + 64 - 85.8 = 27.2$, so $a = 5.2$ cm.

The cosine rule generalises Pythagoras: when $A = 90^{\circ}$, $\cos A = 0$ and you get $a^2 = b^2 + c^2$.

Finding an angle and checking the other possibility

• With sides $a=5,b=7,c=8$, use $\cos A=(b^2+c^2-a^2)/(2bc)=(49+64-25)/112=11/14$. Thus $A=\cos^{-1}(11/14)\approx38.2^{\circ}$.
• In a sine-rule question, $a=10,A=30^{\circ},b=12$ gives $\sin B=b\sin A/a=0.6$. Both $B\approx36.9^{\circ}$ and $143.1^{\circ}$ leave a positive third angle, so two triangles are possible. Use any extra condition, such as B being obtuse, to choose.

Area with an included angle

• Two sides 7 cm and 8 cm include $40^{\circ}$. Use $K=\frac12bc\sin A=\frac12(7)(8)\sin40^{\circ}\approx18.0\text{ cm}^2$.
• The angle must lie between the two sides used. Keep degree mode and full trigonometric values until the final rounding.

You've got it