Question

triangle xyz has an altitude of h. jayden proves the area of \\(\triangle xyz\\) using trigonometry. his work is shown.
image of triangle xyz with altitude h, sides labeled z, x, b, a, and right angle at the foot of the altitude
the area of \\(\triangle xyz\\) is \\(\frac{1}{2}yh\\) by the definition of the area of a triangle. using the sine ratio, \\(\sin(x) = \frac{h}{x}\\). also, \\(x\sin(x) = h\\) is true by the multiplication property of equality. so, by the substitution property of equality, the area of \\(\triangle xyz\\) is \\(\frac{1}{2}xy\sin(x)\\).
where did jayden make his first error in the proof?
\\(\circ\\) a. the sine ratio \\(\sin(x) = \frac{h}{x}\\) is incorrect. it should be \\(\sin(y) = \frac{h}{x}\\).
\\(\circ\\) b. the area of \\(\triangle xyz\\) is incorrect. it should be \\(\frac{1}{2}xh\\).
\\(\circ\\) c. the sine ratio \\(\sin(x) = \frac{h}{x}\\) is incorrect. it should be \\(\sin(z) = \frac{h}{x}\\).
\\(\circ\\) d. the area of \\(\triangle xyz\\) is incorrect. it should be \\(\frac{1}{2}ah\\).

Explanation:

Step1: Check area definition

The area of a triangle is $\frac{1}{2} \times base \times height$. Here, base is $xy$ (side $XZ$) and height is $h$, so area should be $\frac{1}{2}xyh$, not $\frac{1}{2}yh$.

Step2: Verify sine ratio

$\sin(X) = \frac{opposite}{hypotenuse} = \frac{h}{x}$, which is correct. So the first error is in the area formula.

Answer:

B. The area of △XYZ is incorrect. It should be $\frac{1}{2}xyh$.