Question 460·Hard·Math

The hypotenuse is across from the right angle. Since $26$ is twice $13$, scale the $5$-$12$-$13$ Pythagorean triple by $2$. That gives candidate legs of $10$ and $24$ centimeters. They're candidates because the hypotenuse alone doesn't fix the legs.

A right triangle's area is half the product of its legs. Type $\frac{10\cdot24}{2}$ in Desmos; it prints $120$. That matches the given area, so the shorter leg is $10$ centimeters and the longer leg is $24$ centimeters.

From angle $A$, the longer, $24$-centimeter leg is opposite, or across from the angle. The $10$-centimeter leg is adjacent: it touches $A$ and isn't the hypotenuse. Tangent is opposite divided by adjacent, so write:

Reduce the fraction:

The bracket's $\tan A$ is $\frac{12}{5}$. Choice A.

Call the shorter leg $u$ and the longer leg $v$. The area of a right triangle is half the product of its legs, so:

Multiply by $2$:

The Pythagorean theorem says the legs' squares add to the hypotenuse's square:

We know $uv$ too. Squaring $u+v$ brings those two facts together:

Substitute the known values:

Type $\sqrt{26^2+2(240)}$ in Desmos. It prints $34$, so $u+v=34$; lengths can't be negative.

Since the legs add to $34$, the shorter leg is $u=34-v$. Put that into their product equation:

The longer leg must exceed half of $34$, which is $17$. In Desmos, type $x_1(34-x_1)\sim240\{x_1>17\}$: $x_1$ tells Desmos to solve for a value, and $\sim$ tells it to fit the two sides. Under PARAMETERS, Desmos gives $x_1=24$, so $v=24$.

The leg opposite $A$ is $v=24$; the adjacent leg is $u=34-24$. Type $\frac{24}{34-24}$ in Desmos. It prints $2.4$; its fraction button shows $\frac{12}{5}$. So the bracket's $\tan A=\frac{12}{5}$. Choice A.