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The sine and cosine rules and the area formula for triangles with no right angle, then polar coordinates and parametric equations as two other ways of saying where a point is.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
Right triangles were the easy case. In this lesson you get the two rules that open every other triangle: the sine rule, for when you already have a side and the angle opposite it, and the cosine rule, which is Pythagoras with a correction term that vanishes at a right angle. You also get the area formula that needs no height. Then you meet two further ways to describe a position — polar coordinates, which give a distance and a direction, and parametric equations, which say where a moving point is at each instant — and learn to convert between them and the coordinates you already know.
You can solve a right triangle: Pythagoras for the third side, and sine, cosine or tangent for an angle. You also know that the three angles of any triangle add to $180^\circ$. What you have never had is a way in to a triangle with no right angle in it — and most triangles have none.
Included angle: the angle between two named sides. This is the one the cosine rule wants.
Opposite: side $a$ is opposite angle $A$. The sine rule pairs them.
Sine rule (law of sines): $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$.
Cosine rule (law of cosines): $c^2 = a^2 + b^2 - 2ab\cos C$.
Obtuse: bigger than $90^\circ$; its cosine is negative.
Solve a triangle: find every side and angle you were not given.
The sine rule says every side is in the same proportion to the sine of the angle opposite it: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$ Use it whenever you already have a matched side-and-opposite-angle pair, plus one more piece. If $A = 30^\circ$, $B = 90^\circ$ and $a = 5$, then $b = 5 \times \frac{\sin 90^\circ}{\sin 30^\circ} = 10$. The cosine rule $$c^2 = a^2 + b^2 - 2ab\cos C$$ is Pythagoras with a correction term, and the correction is zero exactly when $C = 90^\circ$. Use it when you have two sides and the angle between them (it gives the third side), or all three sides (rearranged, $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$, it gives any angle). Two sides of $5$ and $8$ around $60^\circ$ give $c^2 = 25 + 64 - 40 = 49$, so $c = 7$. Finally, the area of any triangle is $\tfrac{1}{2}ab\sin C$ — half the product of two sides times the sine of the angle between them, which reduces to the familiar half base times height when $C = 90^\circ$.
| $\theta$ | $\sin\theta$ |
|---|---|
| $15^\circ$ | $0.2588$ |
| $30^\circ$ | $0.5000$ |
| $45^\circ$ | $0.7071$ |
| $60^\circ$ | $0.8660$ |
| $75^\circ$ | $0.9659$ |
| $90^\circ$ | $1.0000$ |
| $105^\circ$ | $0.9659$ |
| $120^\circ$ | $0.8660$ |
| $135^\circ$ | $0.7071$ |
| $150^\circ$ | $0.5000$ |
Another way: picture
A scalene triangle with each vertex labelled $A$, $B$, $C$ and each opposite side $a$, $b$, $c$, and a dotted arc drawn at $C$ between the sides $a$ and $b$ to show which angle the cosine rule means by the included one.
Another way: steps
Using the included angle in the sine rule. The sine rule needs a side and the angle opposite it. Pairing a side with the angle beside it gives a confidently wrong answer.
Losing the minus sign for an obtuse angle. $\cos 120^\circ = -\tfrac{1}{2}$, so $-2ab\cos C$ adds. A triangle with sides $3$ and $5$ around $120^\circ$ has third side $7$, longer than $3 + 5 - 1$ would suggest, not shorter.
Forgetting the square root. The cosine rule gives $c^2$. Writing $c = 49$ instead of $c = 7$ is the single most common slip in this topic.
Two sides and the included angle: the cosine rule. $\cos 120^\circ = -\tfrac{1}{2}$.
Obtuse, so the cosine is negative.
$c^2 = 9 + 25 - 2(3)(5)\left(-\tfrac{1}{2}\right) = 9 + 25 + 15 = 49$.
The correction adds, because the angle is obtuse.
$c = 7$.
Take the root.
Area $= \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(6)(10)\sin 30^\circ$.
The angle must be the included one.
$\sin 30^\circ = \tfrac{1}{2}$, so the area is $30 \times \tfrac{1}{2} = 15$.
Three sides, so rearrange: $\cos C = \dfrac{25 + 25 - 64}{2(5)(5)} = \dfrac{-14}{50}$.
That is $-0.28$; a negative cosine means an obtuse angle, about $106^\circ$.
Two sides of a triangle measure $7$ and $8$, and the angle between them is $120^\circ$. How long is the third side?
Answer:
A triangle has sides $7$, $15$ and $13$. Find, in degrees, the angle between the sides of length $7$ and $15$.
Answer:
In a triangle, $A = 60^\circ$, $B = 15^\circ$ and the side $a$ opposite $A$ measures $15$. Find $b$, to one decimal place. Use $\sin 60^\circ = 0.8660$ and $\sin 15^\circ = 0.2588$.
Answer:
A triangle has two sides of length $5$ and $8$ with an angle of $150^\circ$ between them. What is its area?
Answer:
A walker goes $5$ km in a straight line, turns $60^\circ$ to the left, and walks another $16$ km. How far is she from where she started?
Answer:
You can plot a point from its $x$ and $y$ coordinates, and you know $x = r\cos\theta$ and $y = r\sin\theta$ from the unit circle. You can also solve a simple equation for one letter and substitute it into another. Those two skills are the whole of this lesson, used in two directions.
Polar coordinates $(r, \theta)$: how far out, and in which direction.
Pole: the origin, where $r = 0$.
Negative radius: $(-r, \theta)$ means walk backwards, which lands at $(r, \theta + 180^\circ)$.
Parameter: a third letter, usually $t$, that both $x$ and $y$ depend on — often read as time.
Parametric equations: a pair $x = f(t)$, $y = g(t)$.
Eliminating the parameter: getting rid of $t$ to leave a single equation in $x$ and $y$.
Polar coordinates give a distance and a direction instead of an across and an up. Going out: $x = r\cos\theta$, $y = r\sin\theta$, so $(2, 90^\circ)$ is $(0, 2)$. Coming back: $r = \sqrt{x^2 + y^2}$ and $\tan\theta = \frac{y}{x}$, so $(3, 4)$ is $r = 5$ at about $53^\circ$ — with a check of the quadrant, because the inverse tangent alone cannot tell $(3,4)$ from $(-3,-4)$. A negative radius is legal and means the opposite direction: $(-2, 0^\circ)$ is the point $(2, 180^\circ)$. Parametric equations describe a curve by saying where a moving point is at each instant: $x = 2t$, $y = t^2$. To see what curve that is, eliminate the parameter — solve $x = 2t$ for $t = \frac{x}{2}$ and substitute, giving $y = \frac{x^2}{4}$, a parabola. Some eliminations use an identity instead of a substitution: from $x = 3\cos t$, $y = 3\sin t$, squaring and adding gives $x^2 + y^2 = 9$, a circle of radius $3$. The parametric form carries information the equation loses — where the point starts, which way it goes, and how fast.
Another way: picture
A single point drawn twice on the same axes: once with a dashed rectangle showing its across and up, and once with an arrow from the origin labelled $r$ and an arc from the positive $x$-axis labelled $\theta$ — the same place, two descriptions.
Another way: steps
"A point has one pair of polar coordinates." It has infinitely many. $(2, 90^\circ)$, $(2, 450^\circ)$ and $(-2, 270^\circ)$ are all the same point. Cartesian coordinates are unique; polar ones are not.
Trusting $\arctan(y/x)$ blindly. For $(-3, -4)$ the calculator gives $53^\circ$, which points the wrong way entirely. The inverse tangent cannot tell the third quadrant from the first, so you must check the quadrant yourself and add $180^\circ$ when $x$ is negative.
"Eliminating $t$ gives the whole curve." It gives the equation the curve lies on, but the parameter often traces only part of it, and in a particular direction. $x = t^2$, $y = t$ satisfies $x = y^2$, but only the branch with $x \ge 0$ is ever reached.
$r = \sqrt{1 + 1} = \sqrt{2}$.
Pythagoras, and $r$ is taken positive.
$\tan\theta = \frac{1}{-1} = -1$, and the calculator answers $-45^\circ$ — but the point is in the second quadrant, so $\theta = 135^\circ$.
Always check the quadrant.
The second equation is easier: $t = 3 - y$.
Solve whichever is simpler for $t$.
Substituting, $x = 1 + 2(3 - y) = 7 - 2y$, so $y = \frac{7 - x}{2}$: a straight line of slope $-\tfrac{1}{2}$.
Both coordinates linear in $t$ always gives a line.
$\cos 180^\circ = -1$ and $\sin 180^\circ = 0$.
So $x = 4 \times (-1) = -4$ and $y = 4 \times 0 = 0$: the point $(-4, 0)$.
Convert the polar point $(r, \theta) = (5, 90^\circ)$ to Cartesian coordinates.
(x, y)
Convert the Cartesian point $(35, 12)$ to polar form, with $\theta$ in degrees to the nearest whole number.
r = r, theta = t degrees
The curve $x = 3t$, $y = t^2$ is described without $t$ by which equation?
What curve is traced by $x = 2\cos t$, $y = 2\sin t$ for $0 \le t < 2\pi$?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
You know all three sides of a triangle and want one of its angles. Which rule gets you there?
The polar point $(-4, 0^\circ)$ is the same point as $(4, \theta)$. What is $\theta$, in degrees, between $0$ and $360$?
Answer:
You can solve any triangle and move between Cartesian, polar and parametric descriptions. From memory: which rule do you reach for when you know three sides and no angles, and why does $\arctan(y/x)$ alone not settle the polar angle?
8. Your turn: a triangle with sides $5$, $5$ and $8$ — find the angle between the two fives, step 2
20. Your turn: convert $(4, 180^\circ)$ to Cartesian coordinates, step 2