If $\tan(P) = \dfrac{QR}{PR} = \dfrac{3}{4}$, you can set $QR = 3k$ and $PR = 4k$ for some positive number $k$. Then use the Pythagorean theorem to find the hypotenuse $PQ$.

For angle $Q$, which side is opposite and which is the hypotenuse? Use $\sin(Q) = \dfrac{\text{opposite}}{\text{hypotenuse}}$ with the side lengths you found.

The numbers $3$, $4$, and $5$ often appear together in right triangles. Think about how that pattern can help you quickly get the ratio for $\sin(Q)$.

In Desmos, type P = arctan(3/4) to define angle $P$ whose tangent is $3/4$. Make sure Desmos is in degree mode if you want to think in degrees, or use radians consistently.

Because $PQR$ is a right triangle, $Q = 90^\circ - P$ (or $Q = \pi/2 - P$ in radians). In Desmos, enter Q = 90 - P (or Q = pi/2 - P) to define angle $Q$.

In Desmos, type sin(Q) (or sin(Q) = depending on your setup) and look at the numerical value. This value should match one of the answer choices when written as a simplified fraction.

You can also define QR = 3, PR = 4, PQ = sqrt(3^2 + 4^2) and then compute sinQ = PR / PQ. Compare the decimal value of sinQ with the result from the previous step and with the answer choices.

In right triangle $PQR$, $\tan(P)=\dfrac{3}{4}$.

By definition, for angle $P$ in a right triangle:

• $\tan(P)=\dfrac{\text{opposite to }P}{\text{adjacent to }P}$.

Angle $R$ is the right angle, so side $PQ$ is the hypotenuse.

• The side opposite angle $P$ is $QR$.
• The side adjacent to angle $P$ (but not the hypotenuse) is $PR$.

So $\tan(P)=\dfrac{QR}{PR}=\dfrac{3}{4}$. This means we can let:

• $QR = 3k$
• $PR = 4k$ for some positive scale factor $k$.

Now use the Pythagorean theorem to find the hypotenuse $PQ$.

So

This shows the triangle is a scaled $3$–$4$–$5$ right triangle.

For angle $Q$, sine is defined as

• Angle $Q$ is at vertex $Q$, so the side opposite $Q$ is $PR$.
• The hypotenuse is still $PQ$.

Therefore,

The factor $k$ cancels:

So the value of $\sin(Q)$ is $\dfrac{4}{5}$, which corresponds to choice B.