In triangle ABC, with angle A as 90 degrees, angle B as 25 degrees, and side BC opposite angle B measuring 5 inches, we want to find the length of side AC using the sine ratio.
We know that the sine ratio for a given angle in a right triangle can be expressed as:
\[ \sin(B) = \frac{\text{opposite}}{\text{hypotenuse}} \]
In this triangle:
- The opposite side to angle B is BC.
- The hypotenuse is AB.
Since we know BC = 5 inches, we can express the sine ratio as:
\[ \sin(25^\circ) = \frac{BC}{AB} = \frac{5}{AB} \]
We can rearrange this to find AB:
\[ AB = \frac{5}{\sin(25^\circ)} \]
However, we are looking for the length of AC, which is adjacent to angle B. The cosine ratio can be used to find AC:
\[ \cos(B) = \frac{\text{adjacent}}{\text{hypotenuse}} \] \[ \cos(25^\circ) = \frac{AC}{AB} \]
Substituting in the value of AB from before, we have:
\[ AC = AB \cdot \cos(25^\circ) = \left(\frac{5}{\sin(25^\circ)}\right) \cdot \cos(25^\circ) \]
Thus, the correct use of sine or cosine to find the length of AC involves using the cosine ratio. Therefore, while you can express it in terms of the sine of angle B by first finding AB, the direct cosine ratio gives us:
\[ AC = BC \cdot \cot(25^\circ) \]
The sine ratio directly does not solve for AC specifically without additional information, but through sine, we typically end up needing to find AB first as shown.
In summary: To find AC using sine indirectly, you may use: \[ AC = \frac{5 \cdot \cos(25^\circ)}{\sin(25^\circ)} \] Which simplifies to: \[ AC = 5 \cdot \cot(25^\circ) \]
If you need more assistance with this triangle or how to approach it, please let me know!