Question

find the value of x and y.
(image of a right triangle with one leg labeled 42, one leg labeled x, hypotenuse labeled y, and one acute angle labeled 30°)

Explanation:

Step1: Identify triangle type and trigonometric ratios

This is a right - triangle with one angle \(30^{\circ}\), one angle \(90^{\circ}\), so the third angle is \(60^{\circ}\). We can use trigonometric ratios (\(\tan\), \(\sin\), \(\cos\)) or properties of \(30 - 60-90\) triangles. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite \(30^{\circ}\), opposite \(60^{\circ}\), hypotenuse). Also, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)

For angle \(30^{\circ}\), adjacent side \( = 42\), opposite side \(=x\), hypotenuse \(=y\)

Step2: Find \(x\) using \(\tan\)

\(\tan(30^{\circ})=\frac{x}{42}\)
We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\)
So, \(x = 42\times\tan(30^{\circ})=42\times\frac{1}{\sqrt{3}}=\frac{42}{\sqrt{3}}\)
Rationalizing the denominator: \(x=\frac{42\sqrt{3}}{3} = 14\sqrt{3}\)

Step3: Find \(y\) using \(\cos\)

\(\cos(30^{\circ})=\frac{42}{y}\)
We know that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\)
So, \(y=\frac{42}{\cos(30^{\circ})}=\frac{42}{\frac{\sqrt{3}}{2}}=42\times\frac{2}{\sqrt{3}}=\frac{84}{\sqrt{3}}\)
Rationalizing the denominator: \(y = \frac{84\sqrt{3}}{3}=28\sqrt{3}\times2 = 28\times2 = 56\)? Wait, no, wait. Wait, in a \(30 - 60-90\) triangle, the hypotenuse is twice the shorter leg. The shorter leg is opposite \(30^{\circ}\), but wait, in our triangle, the side adjacent to \(30^{\circ}\) is \(42\), the side opposite \(30^{\circ}\) is \(x\), so the shorter leg is \(x\) (since \(30^{\circ}\) is the smallest angle, its opposite side is the shortest). Wait, maybe I made a mistake earlier. Let's re - evaluate.

In a right - triangle, the side opposite \(30^{\circ}\) is the shortest side. Let's use the property of \(30 - 60-90\) triangle: If the side opposite \(30^{\circ}\) is \(a\), then the hypotenuse \(c = 2a\), and the side opposite \(60^{\circ}\) (adjacent to \(30^{\circ}\)) is \(a\sqrt{3}\)

Here, the side adjacent to \(30^{\circ}\) is \(42\), which is opposite \(60^{\circ}\), so \(a\sqrt{3}=42\), where \(a\) is the side opposite \(30^{\circ}\) (i.e., \(x\))
So, \(a=x=\frac{42}{\sqrt{3}} = 14\sqrt{3}\) (which matches our earlier result for \(x\))
And the hypotenuse \(y = 2a=2\times14\sqrt{3}\)? No, wait, no. Wait, if the side opposite \(60^{\circ}\) is \(42\), then \(a\sqrt{3}=42\), so \(a = \frac{42}{\sqrt{3}}=14\sqrt{3}\) (side opposite \(30^{\circ}\), \(x\)), and hypotenuse \(y = 2a=2\times14\sqrt{3}\)? But that contradicts the \(\cos\) approach. Wait, no, \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{42}{y}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(y=\frac{42}{\frac{\sqrt{3}}{2}}=\frac{84}{\sqrt{3}} = 28\sqrt{3}\times2\)? Wait, \(\frac{84}{\sqrt{3}}=\frac{84\sqrt{3}}{3}=28\sqrt{3}\approx48.497\), but if we use the \(30 - 60-90\) triangle ratio correctly:

Let’s define:
- Let the side opposite \(30^{\circ}\) be \(x\) (shorter leg)
- Side opposite \(60^{\circ}\) be \(42\) (longer leg)
- Hypotenuse be \(y\)

We know that \(\frac{\text{longer leg}}{\text{shorter leg}}=\sqrt{3}\), so \(\frac{42}{x}=\sqrt{3}\), so \(x = \frac{42}{\sqrt{3}}=14\sqrt{3}\)

And \(\frac{\text{hypotenuse}}{\text{shorter leg}} = 2\), so \(\frac{y}{x}=2\), so \(y = 2x=2\times14\sqrt{3}=28\sqrt{3}\)? Wait, no, that can't be. Wait, I think I mixed up the adjacent and opposite sides. Let's use trigonometric functions correctly.

\(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{42}\), \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(x=\frac{42}{\sqrt{3}} = 14\sqrt{3}\)

\(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{42}{y}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(y=\frac{42\times2}{\sqrt{3}}=\frac{84}{\sqrt{3}} = 28\sqrt{3}\approx48.497\)

Wait, but if we consider the triangle with angle \(30^{\circ}\), right - angled, the hypotenuse should be longer than either leg. The leg \(x = 14\sqrt{3}\approx24.248\), leg \(42\), hypotenuse \(y = 28\sqrt{3}\approx48.497\), but \(48.497>42\), so that's correct.

Wait, maybe my initial confusion was due to misidentifying the shorter and longer legs. The side opposite \(30^{\circ}\) is the shortest leg (\(x\)), the side opposite \(60^{\circ}\) (adjacent to \(30^{\circ}\)) is the longer leg (\(42\)), and the hypotenuse (\(y\)) is twice the shortest leg.

Since the longer leg \(= \) shorter leg \(\times\sqrt{3}\), so \(42=x\sqrt{3}\), so \(x = \frac{42}{\sqrt{3}}=14\sqrt{3}\)

And hypotenuse \(y = 2x=2\times14\sqrt{3}=28\sqrt{3}\)

Wait, but when we use \(\cos(30^{\circ})=\frac{42}{y}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(y=\frac{42}{\frac{\sqrt{3}}{2}}=\frac{84}{\sqrt{3}} = 28\sqrt{3}\), which matches.

Alternatively, we can use \(\sin(30^{\circ})=\frac{x}{y}\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(x=\frac{y}{2}\)

And from Pythagoras theorem: \(x^{2}+42^{2}=y^{2}\)

Substitute \(x = \frac{y}{2}\) into Pythagoras: \((\frac{y}{2})^{2}+42^{2}=y^{2}\)

\(\frac{y^{2}}{4}+1764=y^{2}\)

\(y^{2}-\frac{y^{2}}{4}=1764\)

\(\frac{3y^{2}}{4}=1764\)

\(y^{2}=1764\times\frac{4}{3}=2352\)

\(y=\sqrt{2352}=\sqrt{49\times48}=\sqrt{49\times16\times3}=7\times4\times\sqrt{3}=28\sqrt{3}\)

And \(x=\frac{y}{2}=14\sqrt{3}\)

Answer:

\(x = 14\sqrt{3}\), \(y = 28\sqrt{3}\) (or if we made a mistake in the \(30 - 60-90\) ratio application earlier, wait, no, let's check with approximate values. \(\sqrt{3}\approx1.732\), \(x = 14\times1.732\approx24.248\), \(y = 28\times1.732\approx48.496\), and \(24.248^{2}+42^{2}\approx587.96 + 1764=2351.96\), and \(48.496^{2}\approx2351.96\), so it's correct. )