A shopkeeper in Chandni Chowk hangs a signboard of mass \(10\,\text{kg}\) from a knot. One string runs from the knot straight across to the wall. The other runs up to a beam of the roof, at \(30°\) to the vertical. Both strings look much alike. Which one is pulled harder, and by how much? The angles between the strings are not right angles, so the ratios of a right triangle alone do not answer at once.
Triangles without a right angle
Sine, cosine and tangent were defined in a right triangle. Forces and velocities form triangles with any angles. Two rules handle every triangle. Name the three sides \(a\), \(b\) and \(c\), and the angles opposite them \(A\), \(B\) and \(C\). Each side and the angle across from it are a pair.
In any triangle, the ratio of a side to the sine of the opposite angle is the same for all three sides. The square of any side equals the sum of the squares of the other two sides, minus a correction. The correction is twice their product times the cosine of the angle between them.
The sine rule has a proof in four lines. Drop a perpendicular of height \(h\) from the corner \(C\) to the side \(c\). It cuts the triangle into two right triangles, and each gives \(h\) in its own way.
\[ h = b\sin A \]
Why. in the right triangle with hypotenuse \(b\), the height \(h\) is across from the angle \(A\).
\[ h = a\sin B \]
Why. in the other right triangle, with hypotenuse \(a\), the height is across from the angle \(B\).
\[ b\sin A = a\sin B \ \Rightarrow\ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} \]
Why. both lines give the same height; divide by \(\sin A\sin B\).
\[ \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \]
Read it. a perpendicular from the corner \(A\) to the side \(a\) gives this pair in the same way.
The cosine rule comes from coordinates. Put the corner \(C\) at the origin and the side \(a\) along the \(x\) axis, so the far corner is at \((a, 0)\). The side \(b\) leaves the origin at the angle \(C\), so its end is at \((b\cos C, b\sin C)\).
\[ c^2 = (a - b\cos C)^2 + (b\sin C)^2 \]
Why. the side \(c\) joins \((a, 0)\) to \((b\cos C, b\sin C)\); use Pythagoras on the differences of the coordinates.
\[ c^2 = a^2 - 2ab\cos C + b^2\cos^2 C + b^2\sin^2 C \]
Why. square out the first bracket.
\[ c^2 = a^2 + b^2 - 2ab\cos C \]
Read it. \(\cos^2 C + \sin^2 C = 1\), so the last two terms are \(b^2\).
In a triangle, the side \(a = 10\,\text{cm}\) is opposite the angle \(A = 30°\), and the angle \(B = 45°\). Find the side \(b\), in cm.
Two sides of a triangle are \(5\,\text{m}\) and \(8\,\text{m}\), and the angle between them is \(60°\). Find the third side, in metres.
The triangle of forces
When three forces keep a body at rest, their vector sum is zero. Draw them head to tail, each arrow starting where the last one ended. The last arrow ends where the first began, so the three arrows close a triangle, the triangle of forces. Its sides are the sizes of the forces. Its angles come from the directions of the forces. Every rule of triangles now applies to the forces.
In the free-body diagram, the three arrows start together at the knot. The angle between two of them there is not the angle inside the triangle. Carrying one arrow to the head of the other turns the angle into its supplement. So an angle of \(\alpha\) at the knot is an angle of \(180° - \alpha\) inside the triangle. Since \(\sin(180° - \alpha) = \sin\alpha\), the sine rule keeps its form with the angles at the knot. That form is Lami's theorem.
A load of mass \(10\,\text{kg}\) hangs from a knot. A horizontal string ties the knot to a wall. A second string ties the knot to the ceiling at \(30°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find the tension \(T_1\) in the ceiling string and \(T_2\) in the horizontal string.
Three forces act on the knot: \(W = 100\,\text{N}\) down, \(T_2\) horizontal and \(T_1\) along the ceiling string. Draw them head to tail: \(W\) down, \(T_2\) across, and \(T_1\) back to the start.
Inside the triangle, the angle between \(W\) and \(T_1\) is \(30°\), and \(T_2\) is across from it. The angle between \(W\) and \(T_2\) is \(90°\), across from \(T_1\). The third angle, \(60°\), is across from \(W\). The sine rule: \(\dfrac{W}{\sin 60°} = \dfrac{T_1}{\sin 90°} = \dfrac{T_2}{\sin 30°}\).
\(\dfrac{100}{0.866} = 115.5\,\text{N}\). So \(T_1 = 115.5 \times 1 = 115.5\,\text{N}\) and \(T_2 = 115.5 \times 0.5 = 57.7\,\text{N}\).
The ceiling string carries \(115.5\,\text{N}\), more than the whole weight, and the wall string \(57.7\,\text{N}\). The cosine rule checks it: \(115.5^2 + 57.7^2 - 2 \times 115.5 \times 57.7\cos 60° = 10\,000 = 100^2\).
The signboard's question is answered. The string to the beam is pulled with \(115.5\,\text{N}\), twice as hard as the string to the wall. The steeper string takes the weight; the level string only holds the knot in place sideways. Moving the beam string closer to the vertical in the sim lowers both tensions.
Three forces \(P\), \(Q\) and \(R\) keep a ring in equilibrium. In Lami's theorem, which angle goes under \(P\)?
The resultant of two vectors
Two forces pull on one point, or a boat moves in a moving river. The sum of two vectors \(P\) and \(Q\) is found by the triangle rule. Draw \(Q\) from the head of \(P\). The arrow from the tail of \(P\) to the head of \(Q\) is the resultant \(R\). Let \(\theta\) be the angle between \(P\) and \(Q\) when their tails meet. Inside the triangle, the angle between them is then \(180° - \theta\).
Apply the cosine rule with that interior angle: \(R^2 = P^2 + Q^2 - 2PQ\cos(180° - \theta)\). Since \(\cos(180° - \theta) = -\cos\theta\), the minus becomes a plus. So \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\). The direction follows from the sine rule. The resultant makes an angle \(\alpha\) with \(P\), where \(\sin\alpha = Q\sin\theta/R\).
The two ends make sense. At \(\theta = 0\) the forces pull together, \(\cos 0 = 1\), and \(R = P + Q\). At \(\theta = 180°\) they pull against each other, \(\cos 180° = -1\), and \(R = |P - Q|\). At \(90°\) the cosine term is zero and \(R = \sqrt{P^2 + Q^2}\), Pythagoras again. Every resultant lies between \(|P - Q|\) and \(P + Q\).
Two SHMs of one frequency along the same line have amplitudes \(3\,\text{cm}\) and \(2\,\text{cm}\), and the second is \(60°\) ahead of the first in phase. Their phasors add like vectors. Find the amplitude of the resultant motion, in cm.
Two forces of \(10\,\text{N}\) each pull on a ring, with \(120°\) between them. What is the size of their resultant?
The cosine rule \(c^2 = a^2 + b^2 - 2ab\cos C\) needs the angle inside the triangle. The resultant formula \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\) needs the angle between the tails. Mixing the two flips the sign. The sine rule has a trap of its own. It gives the sine of an angle, and an angle and its supplement share one sine. If the unknown angle might be larger than \(90°\), check it with the cosine rule or with the angle sum of \(180°\).
Objects in Equilibrium solves the triangle of forces of a knot or a hanging lamp. It writes the sine rule of that triangle as Lami's theorem. Relative Velocity in Two Dimensions finds the size of a relative velocity from the cosine rule when two velocities make any angle. Superposition, Damped and Forced Oscillations adds two SHMs of one frequency as two phasors, with the resultant amplitude from the same rule. The chapters on electric and magnetic fields add the fields of two sources at an angle in the same way.
- Pairing a side with an angle that touches it.Each side goes with the angle across from it. In Lami's theorem, each force goes with the angle between the other two.
- Using the angle between the tails in the cosine rule.The cosine rule needs the interior angle, \(180° - \theta\). With the angle between the tails, use \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\).
- Trusting the sine rule for an angle that may be obtuse.\(\sin\alpha = \sin(180° - \alpha)\). Decide between the two with the cosine rule or the angle sum.