Question

which equation can be used to solve for the measure of angle abc? \\( \tan(x) = \frac{2.4}{10} \\) \\( \tan(x) = \frac{10}{2.4} \\) \\( \sin(x) = \frac{10}{10.3} \\) \\( \sin(x) = \frac{10.3}{10} \\) not drawn to scale triangle with right angle at c, ac = 2.4 cm, bc = 10 cm, ab = 10.3 cm, angle at b is x.

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\tan(x)=\frac{\text{opposite}}{\text{adjacent}}\).
For angle \(x = \angle ABC\), the opposite side to \(\angle ABC\) is \(AC = 2.4\) cm, the adjacent side is not relevant for \(\sin\) calculation here. The hypotenuse of the right - triangle \(\triangle ABC\) is \(AB=10.3\) cm and the side \(BC = 10\) cm is not the hypotenuse.

Step2: Apply the sine formula

Since \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\), and the opposite side to \(\angle ABC\) is \(AC = 2.4\) (incorrect for options A and B as they use \(\tan\) in wrong ratio), for \(\sin(x)\), with opposite \(AC = 2.4\) (not in options) but if we consider the correct sides:
In right - triangle \(\triangle ABC\) with right - angle at \(C\), \(\sin(\angle ABC)=\frac{AC}{AB}\) (wrong as \(AC = 2.4\) and \(AB = 10.3\) not in options). Wait, re - checking:
Wait, no, actually, if we use the sides correctly: \(\sin(x)=\frac{AC}{AB}\) (but \(AC = 2.4\), \(AB=10.3\) is not an option). Wait, no, mistake in step 1.
Wait, correct: In right - triangle \(\triangle ABC\) (\(\angle C = 90^{\circ}\)), for \(\angle ABC=x\), \(\sin(x)=\frac{AC}{AB}\) (but \(AC = 2.4\), \(AB = 10.3\) is not an option). Wait, no, another approach:
By the definition of sine in a right - triangle \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle ABC\), the opposite side is \(AC = 2.4\) (incorrect for options C and D). Wait, no, wait the formula \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\), if we consider the sides:
The hypotenuse \(AB = 10.3\), the side opposite to \(\angle ABC\) is \(AC=2.4\) (not in options). But if we use the formula correctly for the given options:
\(\sin(x)=\frac{AC}{AB}\) (but \(AC = 2.4\), \(AB = 10.3\) is not an option). Wait, no, wait the problem may have a mis - labeling (but assuming the options):
By the formula \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\), if we take the side \(AC = 2.4\) (opposite), \(AB = 10.3\) (hypotenuse) is not an option. But if we check the options:
\(\sin(x)=\frac{10}{10.3}\): No, because \(10\) is \(BC\) (adjacent to \(\angle A\), opposite to \(\angle A\) is \(BC\) but no. Wait, no, another way:
Let's recall that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). In \(\triangle ABC\) (\(\angle C = 90^{\circ}\)), for \(\angle ABC=x\), \(\sin(x)=\frac{AC}{AB}\) (but \(AC = 2.4\), \(AB=10.3\)). But if we check the options:
Option C: \(\sin(x)=\frac{10}{10.3}\) is wrong (as \(10\) is \(BC\), adjacent to \(\angle A\)). Option D: \(\sin(x)=\frac{10.3}{10}\) (incorrect as hypotenuse over side).
Now for \(\tan(x)=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle ABC=x\), opposite is \(AC = 2.4\), adjacent is \(BC = 10\). So \(\tan(x)=\frac{2.4}{10}\) (Option A)

Answer:

\(\tan(x)=\frac{2.4}{10}\) (Option A)