Question

isaiah sketches a model of a skateboard ramp. the model has two surfaces on which to skate, represented by sides ab and ad in the diagram. the steepest side of the model, ab, measures 4 inches. what is the length of the other skating surface, ad?
options:

• ( 2sqrt{2} ) in.
• ( 2sqrt{3} ) in.
• ( 4sqrt{2} ) in.
• ( 4sqrt{3} ) in.

diagram: triangle with a at top, b and d at base ends, c is foot of perpendicular from a to bd, right angle at c. angles: at b: 45°, at a (between ab and ac): 45°, at a (between ac and ad): 60°, at d: 30°.

Explanation:

Step1: Find AC in triangle ABC

In triangle ABC, angle B is 45°, angle C is 90°, so triangle ABC is an isosceles right triangle. AB = 4 inches. Using the sine function: $\sin(45^\circ)=\frac{AC}{AB}$. Since $\sin(45^\circ)=\frac{\sqrt{2}}{2}$, we have $\frac{\sqrt{2}}{2}=\frac{AC}{4}$. Solving for AC: $AC = 4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$? Wait, no, wait. Wait, angle at B is 45°, so in triangle ABC, angle B = 45°, angle C = 90°, so angle A (at A in triangle ABC) is 45°, so AC = BC, and AB is hypotenuse. So $\sin(45^\circ)=\frac{AC}{AB}$, so $AC = AB\sin(45^\circ)=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$? Wait, no, wait, maybe I made a mistake. Wait, no, let's check triangle ACD. In triangle ACD, angle D is 30°, angle C is 90°, so AC is opposite angle D (30°), and AD is hypotenuse. Wait, first, let's confirm AC. In triangle ABC, AB is 4, angle B is 45°, right angle at C. So $\cos(45^\circ)=\frac{BC}{AB}$, $\sin(45^\circ)=\frac{AC}{AB}$. So $AC = AB\sin(45^\circ)=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$? Wait, no, wait, 4 times (√2/2) is 2√2? Wait, 4*(√2/2) = 2√2. Then, in triangle ACD, angle D is 30°, angle C is 90°, so AC is opposite angle D (30°), and AD is hypotenuse. So $\sin(30^\circ)=\frac{AC}{AD}$. Since $\sin(30^\circ)=\frac{1}{2}$, we have $\frac{1}{2}=\frac{AC}{AD}$, so $AD = 2\times AC$. Wait, but earlier we found AC as 2√2? That can't be, because then AD would be 4√2, but that's one of the options. Wait, but let's re-examine. Wait, maybe in triangle ABC, angle at A is 45°, so triangle ABC is isosceles right triangle, so AC = AB * sin(45°) = 4*(√2/2) = 2√2. Then in triangle ACD, angle D is 30°, so sin(30°) = AC / AD => 1/2 = (2√2)/AD => AD = 4√2. Wait, but that's option C: 4√2 in. But wait, let's check again. Wait, maybe I messed up the angle in triangle ACD. Wait, the diagram shows angle at A (between AC and AD) is 60°, so in triangle ACD, angle at A is 60°, angle at D is 30°, angle at C is 90°. So in triangle ACD, angle D is 30°, so the side opposite 30° is AC, and hypotenuse is AD. So $\sin(30^\circ)=\frac{AC}{AD}$, so $AD = \frac{AC}{\sin(30^\circ)}$. But first, find AC from triangle ABC. In triangle ABC, AB = 4, angle B = 45°, right angle at C. So AC = AB * sin(45°) = 4*(√2/2) = 2√2. Then in triangle ACD, sin(30°) = AC / AD => 1/2 = (2√2)/AD => AD = 4√2. Which is option C. Wait, but let's check again. Alternatively, maybe triangle ABC is a 45-45-90 triangle, so AC = AB * cos(45°)? Wait, no, cos(45°) is adjacent over hypotenuse. In triangle ABC, angle at B is 45°, so adjacent side to B is BC, opposite is AC. So sin(45°) = AC / AB, so AC = AB sin(45°) = 4*(√2/2) = 2√2. Then in triangle ACD, angle at D is 30°, so sin(30°) = AC / AD => AD = AC / sin(30°) = (2√2) / (1/2) = 4√2. So that's 4√2 inches, which is option C.

Step2: Calculate AD using triangle ACD

We found AC = 2√2 inches from triangle ABC. In triangle ACD, angle D = 30°, angle C = 90°, so $\sin(30^\circ)=\frac{AC}{AD}$. Since $\sin(30^\circ)=\frac{1}{2}$, we have $\frac{1}{2}=\frac{2\sqrt{2}}{AD}$? Wait, no, wait, that can't be, because then AD would be 4√2, which is option C. Wait, but let's check the angles again. The diagram shows angle at A (between AB and AC) is 45°, between AC and AD is 60°, so angle BAD is 45°+60°=105°, but maybe that's not necessary. Alternatively, maybe I made a mistake in AC. Wait, maybe triangle ABC is a 45-45-90 triangle, so AC = AB * cos(45°)? Wait, cos(45°) is √2/2, so AC = 4*(√2/2) = 2√2. Then in triangle ACD, angle D is 30°, so AD is hypotenuse, AC is opposite 30°, so AD = 2*AC? Wait, no, in a 30-60-90 triangle, the hypotenuse is twice the shorter leg (opposite 30°). So if angle D is 30°, then AC is opposite 30°, so AD (hypotenuse) is twice AC? Wait, no, 30-60-90 triangle: sides are x, x√3, 2x, where x is opposite 30°, x√3 opposite 60°, 2x hypotenuse. So if AC is opposite 30° (angle D), then AC = x, AD = 2x. But we found AC = 2√2, so AD = 4√2, which is option C. So that's correct.

Answer:

$\boldsymbol{4\sqrt{2}}$ in. (Corresponding to the option with $4\sqrt{2}$ in.)