Answer :
To determine the length of the shortest leg of a right triangle where one of the angles is 60 degrees and the longer leg is 15, we can use the properties of a 30-60-90 triangle.
Properties of a 30-60-90 Triangle:
- In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).
- The shortest leg (opposite the 30-degree angle) is \(1\) unit.
- The longer leg (opposite the 60-degree angle) is \(\sqrt{3}\) units.
- The hypotenuse is \(2\) units.
Given:
- The longer leg \( = 15\)
- The one angle opposite the shorter leg is \(60 \text{ degrees}\).
To find the length of the shortest leg, we use the ratio properties of the 30-60-90 triangle. Specifically, the longer leg is \(\sqrt{3}\) times the shortest leg.
Let \( x \) be the length of the shortest leg.
According to the 30-60-90 triangle properties:
[tex]\[ \text{Longer leg} = x \cdot \sqrt{3} \][/tex]
Given that the longer leg is \( 15 \):
[tex]\[ 15 = x \cdot \sqrt{3} \][/tex]
To isolate \( x \):
[tex]\[ x = \frac{15}{\sqrt{3}} \][/tex]
Rationalize the denominator by multiplying the numerator and the denominator by \(\sqrt{3}\):
[tex]\[ x = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3} \][/tex]
Therefore, the length of the shortest leg is:
[tex]\[ \boxed{5 \sqrt{3}} \][/tex]
Properties of a 30-60-90 Triangle:
- In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).
- The shortest leg (opposite the 30-degree angle) is \(1\) unit.
- The longer leg (opposite the 60-degree angle) is \(\sqrt{3}\) units.
- The hypotenuse is \(2\) units.
Given:
- The longer leg \( = 15\)
- The one angle opposite the shorter leg is \(60 \text{ degrees}\).
To find the length of the shortest leg, we use the ratio properties of the 30-60-90 triangle. Specifically, the longer leg is \(\sqrt{3}\) times the shortest leg.
Let \( x \) be the length of the shortest leg.
According to the 30-60-90 triangle properties:
[tex]\[ \text{Longer leg} = x \cdot \sqrt{3} \][/tex]
Given that the longer leg is \( 15 \):
[tex]\[ 15 = x \cdot \sqrt{3} \][/tex]
To isolate \( x \):
[tex]\[ x = \frac{15}{\sqrt{3}} \][/tex]
Rationalize the denominator by multiplying the numerator and the denominator by \(\sqrt{3}\):
[tex]\[ x = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3} \][/tex]
Therefore, the length of the shortest leg is:
[tex]\[ \boxed{5 \sqrt{3}} \][/tex]
