To solve for the lengths [tex]\(xy\)[/tex] and [tex]\(yz\)[/tex] in the triangle [tex]\(\Delta xyz\)[/tex] where [tex]\(\angle y = 90^\circ\)[/tex], [tex]\(\angle z = a^\circ\)[/tex], and [tex]\(\angle x = (a + 30)^\circ\)[/tex], we follow these steps:
1. Sum of Angles in a Triangle:
- Given that the sum of angles in any triangle is [tex]\(180^\circ\)[/tex]:
[tex]\[
\angle x + \angle y + \angle z = 180^\circ
\][/tex]
- Substitute the given angles:
[tex]\[
(a + 30)^\circ + 90^\circ + a^\circ = 180^\circ
\][/tex]
- Combine like terms:
[tex]\[
2a + 120 = 180
\][/tex]
- Solve for [tex]\(a\)[/tex]:
[tex]\[
2a = 60 \implies a = 30
\][/tex]
Therefore,
[tex]\[
\angle z = 30^\circ \quad \text{and} \quad \angle x = 30 + 30 = 60^\circ
\][/tex]
2. Given Side Length:
- The length of the hypotenuse [tex]\(xz\)[/tex] is given as 24.
3. Using Trigonometric Functions in a Right Triangle:
- Since we are given a right triangle with [tex]\(\angle y = 90^\circ\)[/tex], we can use the sine and cosine functions to find the other two sides.
4. Finding [tex]\(yz\)[/tex] (adjacent to [tex]\(\angle x = 60^\circ\)[/tex]):
- The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse:
[tex]\[
\cos(60^\circ) = \frac{yz}{xz}
\][/tex]
- Given [tex]\(\cos(60^\circ) = 0.5\)[/tex] and [tex]\(xz = 24\)[/tex]:
[tex]\[
0.5 = \frac{yz}{24}
\][/tex]
- Solve for [tex]\(yz\)[/tex]:
[tex]\[
yz = 24 \times 0.5 = 12
\][/tex]
5. Finding [tex]\(xy\)[/tex] (opposite to [tex]\(\angle x = 60^\circ\)[/tex]):
- The sine of an angle in a right triangle is the ratio of the opposite side to the hypotenuse:
[tex]\[
\sin(60^\circ) = \frac{xy}{xz}
\][/tex]
- Given [tex]\(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)[/tex] and [tex]\(xz = 24\)[/tex]:
[tex]\[
\frac{\sqrt{3}}{2} = \frac{xy}{24}
\][/tex]
- Solve for [tex]\(xy\)[/tex]:
[tex]\[
xy = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3} \approx 20.7846
\][/tex]
Thus, the lengths of the sides are:
- [tex]\(yz = 12\)[/tex]
- [tex]\(xy = 12\sqrt{3} \approx 20.7846\)[/tex]
