Question 71·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

Let $b$ be the number of degrees in $\angle ABC$. Since $D$ lies on $BC$, ray $BD$ points along ray $BC$, so the angle at $B$ is unchanged: $m\angle ABD=b^\circ$.

In isosceles triangle $ABD$, $AB=AD$. Equal sides face equal angles in the same triangle. Those sides face $\angle ADB$ and $\angle ABD$, so $m\angle ADB=b^\circ$.

Triangle $ABC$ has angles $b^\circ$ at $B$ and $t^\circ$ at $C$. Its three angles add to $180^\circ$, so subtract those two to get the whole angle at $A$:

To bisect an angle means to split it into two equal angles. Divide the whole angle at $A$ by $2$ to get the part inside triangle $ACD$:

Ray $DB$ extends side $DC$ of triangle $ACD$. By the exterior-angle theorem, the outside angle $\angle ADB$ equals the two far angles, $\angle CAD$ and $\angle ACD$. Include the angle at $C$ when you write the equation:

Try $t=30$; a correct formula must work for every valid test angle. Type $t=30$ and $x_1\sim\frac{180-x_1-t}{2}+t$. Here $x_1$ stands for $b$, and $\sim$ asks Desmos to fit its value. Desmos reports $x_1=70$ under PARAMETERS. The last line, $180-x_1-t$, prints $80$ for the angle at $A$, so this test gives a valid triangle.

Type the four choices in order. Desmos prints $70$, $50$, $75$, and $105$. Only $60+\frac{t}{3}$ gives the $70^\circ$ angle at $B$ when $t=30$. So the measure of $\angle ABC$ is $\left(60+\frac{t}{3}\right)^\circ$. Choice A.

The Desmos test identifies the choice. To prove it works for every valid $t$, keep $t$ as a letter in the exterior-angle equation. Starting with $b=\frac{180-b-t}{2}+t$, multiply both sides by $2$:

Combine $-t$ and $2t$:

Add $b$ to both sides:

Divide both sides by $3$ to find the angle at $B$:

Split the fraction across the sum, using $180\div3=60$: $b=60+\frac{t}{3}$. Since $b$ is the measure of $\angle ABC$ in degrees, its measure is $\left(60+\frac{t}{3}\right)^\circ$. Choice A.