sat suite question viewer

Choice B is correct. It's given that in $△ABC$, $\angle B$ is a right angle. Therefore, $△ABC$ is a right triangle, and $\overline{AC}$ is the hypotenuse of the triangle. It's also given that $\cos A=\frac{3}{5}$. Since the cosine of an acute angle in a right triangle is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse, the ratio of the length of $\overline{AB}$ to the length of $\overline{AC}$ is $3$ to $5$. It follows that the length of $\overline{AB}$ can be represented as $3a$ and the length of $\overline{AC}$ can be represented as $5a$, where $a$ is a constant. The Pythagorean theorem states that the sum of the squares of the length of the legs of a right triangle is equal to the square of the length of its hypotenuse, so it follows that $A{B}^2+B{C}^2=A{C}^2$. Substituting $3a$ for $AB$ and $5a$ for $AC$ in this equation yields ${\left(3a\right)}^2+B{C}^2={\left(5a\right)}^2$, or $9a^2+B{C}^2=25a^2$. Subtracting $9a^2$ from both sides of this equation yields $B{C}^2=16a^2$, or $BC=4a$. It follows that the ratio of the length of $\overline{AB}$ to the length of $\overline{BC}$ is $3$ to $4$. Let $x$ represent the length, in millimeters, of $\overline{AB}$. It's given that the length of $\overline{BC}$ is $136$ millimeters. Since the ratio of the length of $\overline{AB}$ to the length of $\overline{BC}$ is $3$ to $4$, $\frac{x}{136}=\frac{3}{4}$. Multiplying both sides of this equation by $136$ yields $x=\frac{3\left(136\right)}{4}$, or $x=102$. Therefore, the length of $\overline{AB}$ is $102$ millimeters.

Choice A is incorrect. This is the scale factor by which the $3$ to $4$ to $5$ ratio is multiplied that results in the side lengths of $△ABC$.

Choice C is incorrect. This is the length of $\overline{BC}$, not the length of $\overline{AB}$.

Choice D is incorrect. This is the length of $\overline{AC}$, not the length of $\overline{AB}$.