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Dilations and the similarity criteria, and the sine, cosine and tangent ratios that similarity makes possible.
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.
In this lesson you use dilations to define similarity: what a scale factor does to a point, to a length, to a perimeter and to an area, and how the AA criterion lets you prove two triangles similar from angles alone. You then use similarity to find missing sides in proportion. That work pays off immediately in trigonometry, because similar triangles are the reason the sine, cosine and tangent of an angle are properties of the angle and not of the triangle you happened to draw. You define all three from a right triangle, compute them exactly in whole-number triples, and see why the sine of an angle is the cosine of its complement.
You can plot a point and multiply its coordinates, you have met congruent figures and the rigid motions that produce them, and you can solve a proportion such as $\frac{x}{9} = \frac{12}{4}$. Similarity is congruence with one extra move allowed — resizing — and every missing length here comes out of a proportion you can already solve.
Dilation: the transformation that fixes one point, the centre, and multiplies every distance from it by a scale factor $k$.
Similar: related by a dilation together with rigid motions. Written $\triangle ABC \sim \triangle DEF$, and the order of the letters names which vertex matches which.
Corresponding sides: the pair occupying the same position in the two figures — matched by the angles they sit between, never by size alone.
Enlargement: a dilation with $k > 1$. Reduction: one with $k < 1$.
AA criterion: two pairs of equal angles are enough to prove two triangles similar.
A dilation with centre the origin and scale factor $k$ sends $(x, y)$ to $(kx, ky)$: $(2, 3)$ with $k = 3$ becomes $(6, 9)$, and reading the factor back off is a division, $6 \div 2 = 3$. It multiplies every length by $k$ and changes no angle at all. Two figures are similar when a dilation plus rigid motions carries one onto the other, so corresponding angles are equal and corresponding sides share a single ratio. Triangles with sides $3, 4, 6$ and $9, 12, x$ have factor $9 \div 3 = 3$, so $x = 18$. The AA criterion shortcuts all of this: angles $50^\circ$ and $60^\circ$ in one triangle and $50^\circ$ and $60^\circ$ in another force $70^\circ$ in both, so they are similar. Because a dilation by $k$ scales every length by $k$, the perimeter scales by $k$ and the area by $k^2$.
Another way: picture
A small triangle and, drawn from the same centre point, rays through each of its vertices carrying a second triangle three times as far out. Same angles, every side three times as long.
Another way: story
Similar is a photo enlargement. Every length is multiplied by the same number and nothing is bent, which is why an enlarged photo still looks like the person and not like a funhouse mirror.
Adding instead of multiplying. A side of $4$ becoming $10$ is a factor of $2.5$, not "plus $6$". Test any factor you find on a second pair of sides: an additive rule will fail there, and a genuine scale factor will not.
Scaling area by the scale factor. A dilation by $3$ multiplies every length by $3$ and every area by $9$, because an area is two lengths multiplied together. Lengths, areas and volumes scale by $k$, $k^2$ and $k^3$.
Waiting for a side before claiming similarity. AA needs no sides at all. Two pairs of equal angles force the third pair, because the angles of each triangle add to $180^\circ$, and three equal angles is what having the same shape means.
A dilation from the origin sends $(2, 3)$ to $(8, 12)$. Divide: $8 \div 2 = 4$, and check on the other coordinate, $12 \div 3 = 4$.
One factor has to work for every pair, or it is not a dilation.
A triangle with sides $3, 4, 6$ is dilated by that same $4$, so its sides become $12, 16, 24$.
Multiply each side; do not add $6$ to each.
$\triangle ABC$ has $50^\circ$ and $60^\circ$, so its third angle is $180 - 50 - 60 = 70^\circ$.
Fill in the missing angle before comparing anything.
$\triangle DEF$ has $60^\circ$ and $70^\circ$, so its third is $50^\circ$: the same three angles.
Similar by AA — and no side length was needed.
Two matching pairs already forced the third.
Scale factor from the first pair: $15 \div 5 = 3$, checked by $21 \div 7 = 3$.
$x = 9 \times 3 = 27$.
A dilation with centre the origin sends $(5, 3)$ to $(20, 12)$. What is the scale factor?
answer
Triangle $ABC$ has sides $7$, $6$ and $10$. Triangle $PQR$ is similar to it, with the sides corresponding in that order, and its first two sides are $28$ and $24$. How long is its third side?
Answer:
A triangle is dilated by a scale factor of $4$. What happens to its area?
Triangle $ABC$ has angles of $62^\circ$ and $50^\circ$. Triangle $DEF$ has angles of $50^\circ$ and $72^\circ$. Are the two triangles similar?
You have just proved that two triangles with equal angles are similar, and you know the Pythagorean theorem and how to simplify a fraction. That is the whole toolkit: the ratios below exist only because similarity guarantees they do not depend on how big the triangle is.
Hypotenuse: the side opposite the right angle, always the longest.
Opposite and adjacent: named relative to the acute angle you have chosen. The opposite is the leg the angle faces; the adjacent is the other leg, the one the angle is built on. Change angles and these two swap.
Sine, cosine, tangent: written $\sin\theta$, $\cos\theta$, $\tan\theta$; opposite over hypotenuse, adjacent over hypotenuse, opposite over adjacent.
SOH CAH TOA: the three definitions as one word each.
Complement: what an angle needs to reach $90^\circ$. The two acute angles of a right triangle are complements.
Pick one acute angle $\theta$ of a right triangle. The three sides then have names: the hypotenuse faces the right angle, the opposite faces $\theta$, and the adjacent is the remaining leg. Then $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ — SOH CAH TOA. In a $3, 4, 5$ triangle the angle facing the $3$ has $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ and $\tan\theta = \frac{3}{4}$. These are properties of the angle, not of the triangle: any two right triangles with the same acute angle are similar by AA, so their sides share one scale factor, which cancels out of every ratio. The two acute angles are complements, and the side opposite one is adjacent to the other, so $\sin\theta = \cos(90^\circ - \theta)$: $\sin 30^\circ = \cos 60^\circ$ and $\sin 35^\circ = \cos 55^\circ$.
Another way: picture
One right triangle drawn twice the size of another beside it, the same acute angle marked in both. The sides are all different, but opposite over hypotenuse reduces to the same fraction in each.
Another way: steps
Mark the angle. Label the hypotenuse, then the opposite, then the adjacent. Ask which two of those three the question mentions. That pair names the ratio, and only then do you write a fraction.
Fixing the labels to the triangle instead of to the angle. "Opposite" and "adjacent" belong to the angle you picked, not to the picture. In a $3, 4, 5$ triangle the side of length $3$ is opposite one acute angle and adjacent to the other, so it appears in $\sin$ of one and $\cos$ of the other.
Putting the hypotenuse into a tangent. Tangent is the two legs only. If your tangent came out less than $1$ every single time, you probably divided by the hypotenuse.
Thinking a bigger triangle has bigger ratios. Double every side and the ratios do not move, because the two triangles are similar. That is precisely why $\sin 30^\circ$ can be quoted as a number at all, without saying which triangle it came from.
Take the acute angle $\theta$ facing the side of length $3$. Opposite $= 3$, hypotenuse $= 5$, adjacent $= 4$.
Label first, compute second.
$\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$, $\tan\theta = \frac{3}{4}$.
Both the sine and the cosine are under $1$; the tangent need not be.
In the $6, 8, 10$ triangle the same angle gives $\frac{6}{10} = \frac{3}{5}$ again.
Similar triangles, so the scale factor cancels.
The other acute angle is $90 - 35 = 55^\circ$, because the two acute angles are complements.
The side opposite $35^\circ$ is the side adjacent to $55^\circ$, and both ratios divide it by the same hypotenuse.
One length, two names.
So $\sin 35^\circ = \cos 55^\circ$.
For that angle the opposite is $5$ and the adjacent is $12$; the $13$ takes no part.
$\tan\theta = \frac{5}{12}$.
A right triangle has legs $18$ and $80$ and hypotenuse $82$. Let $\theta$ be the acute angle opposite the side of length $18$. What is $\sin\theta$, as an exact fraction?
Answer:
The same right triangle has legs $36$ and $48$ and hypotenuse $60$. Let $\phi$ be the acute angle opposite the side of length $48$. What is $\tan\phi$, as an exact fraction?
Answer:
In a right triangle one acute angle measures $54^\circ$. There is an angle $\theta$ for which $\sin 54^\circ = \cos\theta$. How many degrees is $\theta$?
Answer:
A right triangle has legs $10$ and $24$ and hypotenuse $26$. Let $\theta$ be the acute angle opposite the side of length $10$. What is $\sin\theta$, as an exact fraction?
Answer:
Two loading ramps are built beside each other. The first rises $5$ cm over a horizontal run of $12$ cm. The second rises $16$ cm over a horizontal run of $30$ cm. Which ramp is steeper?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A dilation with centre the origin sends $(5, 4)$ to $(10, 8)$. What is the scale factor?
answer
The same right triangle has legs $3$ and $4$ and hypotenuse $5$. Let $\phi$ be the acute angle opposite the side of length $4$. What is $\tan\phi$, as an exact fraction?
Answer:
You can find a scale factor, prove triangles similar by AA, and write the three ratios for a chosen acute angle. Explain why enlarging a right triangle leaves its sine unchanged, and say which ratio never uses the hypotenuse.
8. Your turn: triangles with sides $5, 7, 9$ and $15, 21, x$, step 2
19. Your turn: a right triangle with legs $5$ and $12$ and hypotenuse $13$; find $\tan$ of the angle facing the $5$, step 2