Geometry: Similarity & Right-Triangle Trigonometry

How to do it:

1. Check AA: two pairs of congruent angles force similar triangles.

2. Write a scale factor $k$ as corresponding side over corresponding side.

3. Set up a proportion for a missing length. Keep vertices in matching order.

Worked example: A 6 ft person casts an 8 ft shadow. A flagpole casts a 20 ft shadow at the same time. $\tfrac{h}{20} = \tfrac{6}{8}\text{,}$ so $h = 15$ ft.

Check: match vertices in order. Mixing corresponding sides is the usual error. Review Grade 8 similarity.

How to do it:

1. Sketch the right triangle and mark the acute angle $\theta\text{.}$

2. Label opposite (across from $\theta\text{),}$ adjacent (next to $\theta\text{,}$ not the hypotenuse), and hypotenuse (longest side).

3. Write $\sin\theta=\tfrac{\text{opp}}{\text{hyp}}\text{,}$ $\cos\theta=\tfrac{\text{adj}}{\text{hyp}}\text{,}$ $\tan\theta=\tfrac{\text{opp}}{\text{adj}}\text{.}$

Worked example: Opposite 5, hypotenuse 13: $\sin\theta = 5/13\text{.}$ Adjacent 12: $\cos\theta=12/13\text{.}$ Open Sine (SOH).

Check: hypotenuse is opposite the right angle. $\tan\theta = \sin\theta/\cos\theta$ when $\cos\theta \neq 0\text{.}$

How to do it:

1. Sketch and mark the right angle.

2. Choose sin, cos, or tan from the sides you know and want.

3. Use the Pythagorean Theorem as a check: $a^2+b^2=c^2\text{.}$ See Pythagorean theorem.

Worked example: A 30 ft ladder leans at $70°$ to the ground. Height $= 30\sin 70° \approx 28.2$ ft.

Check: an angle of elevation from the ground uses the same triangle as the depression from the top. If the calculator is in radian mode, $70$ is not $70°\text{.}$

How to do it:

1. A dilation with center $O$ and scale factor $k$ sends each point $P$ to the point on ray $OP$ at distance $k\cdot OP\text{.}$ Angles do not change, which is why the image is similar to the original.

2. Every length multiplies by $k\text{,}$ so the perimeter multiplies by $k\text{.}$

3. Area multiplies by $k^2\text{,}$ because both dimensions scale by $k\text{.}$

Worked example: a $3$-$4$-$5$ triangle (perimeter $12\text{,}$ area $6\text{)}$ dilated by $k=3$ becomes $9$-$12$-$15\text{:}$ perimeter $36=12\cdot 3$ and area $54=6\cdot 9\text{.}$ A dilation by $k=\tfrac{1}{2}$ halves every length and quarters the area.

Check: $k>1$ enlarges, $0<k<1$ shrinks, and $k=1$ changes nothing. A photo scaled to $150\%$ has $2.25$ times the area, not $1.5$ times. Replay similar triangles.

How to do it:

1. $45°$-$45°$-$90°\text{:}$ the legs are equal and the hypotenuse is $\text{leg}\cdot\sqrt{2}\text{.}$

2. $30°$-$60°$-$90°\text{:}$ the hypotenuse is twice the short leg, and the long leg is $\text{short leg}\cdot\sqrt{3}\text{.}$ The short leg sits opposite the $30°$ angle.

3. Work backwards by dividing: hypotenuse $\div\sqrt{2}$ gives a $45°$ leg; hypotenuse $\div 2$ gives the short leg.

Worked example: legs of $5$ give hypotenuse $5\sqrt{2}\approx 7.07\text{.}$ A $30°$-$60°$-$90°$ triangle with short leg $4$ has long leg $4\sqrt{3}\approx 6.93$ and hypotenuse $8\text{.}$ With hypotenuse $10\text{,}$ the short leg is $5$ and the long leg $5\sqrt{3}\text{.}$

Check: Pythagoras agrees: $5^2+5^2=50=(5\sqrt{2})^2\text{.}$ These triangles give the exact values $\sin 30°=\tfrac{1}{2}\text{,}$ $\cos 45°=\tfrac{\sqrt{2}}{2}\text{,}$ $\tan 60°=\sqrt{3}\text{.}$ See special right triangles.

How to do it:

1. Build the ratio from the two sides you know.

2. Apply the matching inverse: $\theta=\sin^{-1}\text{,}$ $\cos^{-1}\text{,}$ or $\tan^{-1}$ of that ratio.

3. Make sure the calculator is in degree mode, and expect an answer between $0°$ and $90°\text{.}$

Worked example: opposite $7\text{,}$ adjacent $24\text{:}$ $\theta=\tan^{-1}(7/24)\approx 16.3°\text{.}$ A ramp that rises $1$ for every $12$ of run: $\theta=\tan^{-1}(1/12)\approx 4.8°\text{.}$ A $30$ ft ladder reaching $28.2$ ft up a wall: $\theta=\sin^{-1}(28.2/30)\approx 70°\text{.}$

Check: $\sin^{-1}$ and $\cos^{-1}$ only accept inputs from $-1$ to $1\text{.}$ If the calculator errors on $\sin^{-1}(1.2)\text{,}$ the side you called opposite is longer than the hypotenuse, so relabel. $\tan^{-1}$ accepts any number.

How to do it:

1. The two acute angles of a right triangle add to $90°\text{.}$

2. The side opposite one acute angle is adjacent to the other, so $\sin\theta=\cos(90°-\theta)$ and $\cos\theta=\sin(90°-\theta)\text{.}$

3. Use the swap to read one ratio from the other without recomputing.

Worked example: $\sin 30°=\tfrac{1}{2}$ and $\cos 60°=\tfrac{1}{2}\text{.}$ In a $3$-$4$-$5$ triangle, if $\sin\theta=0.6$ then $\cos\theta=0.8\text{,}$ so $\sin(90°-\theta)=0.8$ and $\cos(90°-\theta)=0.6\text{.}$

Check: "cosine" means the sine of the complement. If $\sin\theta=\cos\theta\text{,}$ the two acute angles are equal, so $\theta=45°\text{.}$

How to do it:

1. Elevation is measured up from the horizontal; depression is measured down from it. When the two horizontals are parallel, the two angles are equal (alternate interior angles).

2. Draw the right triangle with the horizontal as one leg.

3. Add the observer's eye height when the story gives one.

Worked example: from $50$ m away, the angle of elevation to the top of a tower is $38°\text{,}$ measured from eyes $1.6$ m above the ground. Height above eye level $=50\tan 38°\approx 39.1$ m, so the tower is about $40.7$ m tall. From the top of a $120$ ft cliff, the angle of depression to a boat is $25°\text{:}$ the boat is $120/\tan 25°\approx 257$ ft from the base.

Check: an angle under $45°$ means the height is less than the horizontal distance, so $39$ m for $50$ m across makes sense. For the boat, a depression of $25°$ means it is farther out than the cliff is tall.