Right Triangles and Trigonometry Pattern - Ratio Conversion

This pattern gives you one trig ratio in a right triangle and asks you to find a different trig ratio — often for the other acute angle, or the same angle but a different function. The core technique: use the given ratio to reconstruct the triangle's sides, then compute the requested ratio.

The Core Method: Build the Triangle

When you're given something like $\sin Q = \dfrac{7}{25}$, this tells you: - The side opposite angle $Q$ has length $7$ (or $7k$) - The hypotenuse has length $25$ (or $25k$) - The third side (adjacent to $Q$) is found via the Pythagorean theorem: $\sqrt{25^2 - 7^2} = \sqrt{576} = 24$

Now you have all three sides ($7$, $24$, $25$) and can compute any trig ratio for any angle.

Converting Between Angles: $\sin(Q)$ to $\tan(R)$

In triangle $PQR$, angle $P$ is a right angle. If $\sin Q = \dfrac{7}{25}$, what is the value of $\tan R$?

Step 1: Build the triangle.
$\sin Q = \dfrac{\text{opposite } Q}{\text{hypotenuse}} = \dfrac{PR}{QR} = \dfrac{7}{25}$.
Missing side: $PQ = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$.
Sides: $PR = 7$, $PQ = 24$, $QR = 25$.

Step 2: Compute the requested ratio.
$\tan R = \dfrac{\text{opposite } R}{\text{adjacent to } R} = \dfrac{PQ}{PR} = \dfrac{24}{7}$.
The answer is $\dfrac{24}{7} \approx 3.429$.

Why the switch works: In a right triangle, the side opposite one acute angle is adjacent to the other. So $\sin Q$ and $\cos R$ use the same sides (just in a different role), and $\tan R$ flips what $\tan Q$ uses.

Converting Between Functions: $\sin(Q)$ to $\sin(R)$

In triangle $PQR$, $\angle P = 90°$ and $\sin Q = \dfrac{14}{50}$. What is $\sin R$?

Simplify: $\sin Q = \dfrac{14}{50} = \dfrac{7}{25}$.
Build the triangle: opposite to $Q = 7$, hypotenuse $= 25$, adjacent to $Q = 24$.
$\sin R = \dfrac{\text{opposite } R}{\text{hypotenuse}} = \dfrac{PQ}{QR} = \dfrac{24}{25}$.
The answer is $\dfrac{24}{25} = 0.96$.

Notice: $\sin R = \cos Q$, because $Q + R = 90°$ (the cofunction identity from pattern 4). This gives a shortcut: $\sin R = \cos Q = \dfrac{\text{adjacent to } Q}{\text{hypotenuse}} = \dfrac{24}{25}$.

Similar Triangles: Trig Ratios Don't Scale

When two triangles are similar, their trig ratios are identical — the scale factor cancels out. This means information about perimeters or areas being different is irrelevant to finding trig ratios.

$\triangle PQR \sim \triangle STU$, where $\angle R$ and $\angle U$ are right angles. $P$ corresponds to $S$. The perimeter of $\triangle STU$ is $1.5$ times the perimeter of $\triangle PQR$. If $\cos S = \dfrac{8}{17}$, what is $\tan P$?

Since $P$ corresponds to $S$: $\angle P = \angle S$.
The perimeter ratio ($1.5\times$) is irrelevant for trig ratios.
From $\cos S = \dfrac{8}{17}$: adjacent to $S = 8$, hypotenuse $= 17$, opposite $= \sqrt{17^2 - 8^2} = \sqrt{225} = 15$.
$\tan P = \tan S = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{15}{8}$.
The answer is $\dfrac{15}{8} = 1.875$.

$\triangle PQR \sim \triangle TUV$, with $P \leftrightarrow T$, $R \leftrightarrow V$. Right angles at $R$ and $V$. If $\sin P = \dfrac{4}{5}$, what is $\sin U$?

$Q$ corresponds to $U$ (the remaining vertices).
Build the triangle: $\sin P = \dfrac{4}{5}$, so opposite $= 4$, hyp $= 5$, adjacent $= 3$ (a $3$-$4$-$5$ triple).
$\sin U = \sin Q = \dfrac{\text{opposite } Q}{\text{hypotenuse}} = \dfrac{3}{5}$.
The answer is $\dfrac{3}{5} = 0.6$.

Gotcha: $\sin U \neq \sin P$. $U$ corresponds to $Q$ (not $P$), so you need the trig ratio for the other acute angle.

Area Scaling and Trig Ratios

$\triangle STU$ has area $9$ times the area of $\triangle PQR$. The triangles are similar with $P \leftrightarrow S$. If $\cos P = \dfrac{15}{17}$, find $\tan S$.

Area scales as $k^2$. If area ratio is $9$, then $k = 3$ (the linear scale factor). But this doesn't affect trig ratios at all.
$P$ corresponds to $S$: $\cos S = \cos P = \dfrac{15}{17}$.
Build: adjacent $= 15$, hyp $= 17$, opposite $= \sqrt{289 - 225} = 8$.
$\tan S = \dfrac{8}{15}$.
The answer is $\dfrac{8}{15} \approx 0.533$.

Key insight: When problems mention area ratios or perimeter ratios with similar triangles, they're testing whether you'll waste time on the scale factor. Trig ratios are scale-independent.

Summary of the Method

1. Extract the ratio. Write $\sin$, $\cos$, or $\tan$ as $\dfrac{a}{b}$ and identify which sides $a$ and $b$ represent (opposite, adjacent, hypotenuse)
2. Find the missing side. Use $a^2 + b^2 = c^2$ (add for legs → hypotenuse, subtract for hypotenuse minus leg)
3. Identify which angle the question asks about. Is it the same angle, the other acute angle, or a corresponding angle in a similar triangle?
4. Compute the requested ratio. Apply SOHCAHTOA for the correct angle

What to Do on Test Day

• Build the triangle from the ratio. Given $\sin \theta = \dfrac{a}{c}$, the three sides are $a$, $\sqrt{c^2 - a^2}$, and $c$. Recognize Pythagorean triples: $3$-$4$-$5$, $5$-$12$-$13$, $7$-$24$-$25$, $8$-$15$-$17$
• Simplify fractions first. If $\sin Q = \dfrac{14}{50}$, simplify to $\dfrac{7}{25}$ before finding the missing side
• Label all three sides with "opposite," "adjacent," and "hypotenuse" for both acute angles. What is opposite one angle is adjacent to the other
• Trig ratios don't change with scaling. Perimeter ratios, area ratios, and side length differences between similar triangles are irrelevant for trig ratio questions
• Cofunction shortcut: $\sin A = \cos B$ when $A + B = 90°$. If asked for $\sin R$ and you know $\cos Q$, just use the cofunction identity if $Q + R = 90°$
• Watch the correspondence. In $\triangle PQR \sim \triangle STU$: $P \leftrightarrow S$, $Q \leftrightarrow T$, $R \leftrightarrow U$. If asked for $\sin U$, that's $\sin R$, not $\sin P$
• These are all student-produced response (SPR). Double-check your arithmetic — there are no answer choices to guide you. Common accepted formats: fractions like $\dfrac{24}{7}$ or decimals like $3.429$