Question

plane unit 6 (#82)
derive a formula that can be used to find the missing side lengths of any 30°-60°-90° triangle.
to find missing sides in the special right triangle (30°-60°-90°), i will use the formulas:
notes: identify the sides before you start.
remember to rationalize the denominator if needed. answers must be in simplest form.
1.
2.
3.
4.
5.
6.
7.
8.
9.
how do you find missing side lengths in a 30°-60°-90° triangle?
how are the angles and sides of special right triangles related to each other?

Explanation:

Identify the 30-60-90 ratio rules

In any \(30^\circ\)-\(60^\circ\)-\(90^\circ\) right triangle, the side lengths are related as follows:

• Short Leg (\(SL\), opposite the \(30^\circ\) angle): \(x\)
• Long Leg (\(LL\), opposite the \(60^\circ\) angle): \(x\sqrt{3}\)
• Hypotenuse (\(hyp\), opposite the \(90^\circ\) angle): \(2x\)

Solve questions 1 through 3

• Question 1: Given short leg \(SL = 4\).
• \(x = 4\)
• Long leg \(n = SL \cdot \sqrt{3} = 4\sqrt{3}\)
• Hypotenuse \(m = 2 \cdot SL = 2 \cdot 4 = 8\)
• Question 2: Given short leg \(SL = 6\).
• \(x = 6\)
• Hypotenuse \(x = 2 \cdot SL = 12\)
• Long leg \(y = SL \cdot \sqrt{3} = 6\sqrt{3}\)
• Question 3: Given short leg \(SL = \sqrt{3}\).
• \(x = \sqrt{3}\)
• Hypotenuse \(a = 2 \cdot SL = 2\sqrt{3}\)
• Long leg \(b = SL \cdot \sqrt{3} = \sqrt{3} \cdot \sqrt{3} = 3\)

Solve questions 4 through 6

• Question 4: Given hypotenuse \(hyp = 6\).
• \(2x = 6 \implies x = 3\) (short leg)
• Short leg \(b = 3\)
• Long leg \(a = 3\sqrt{3}\)
• Question 5: Given hypotenuse \(hyp = 2\).
• \(2x = 2 \implies x = 1\) (short leg)
• Short leg \(y = 1\)
• Long leg \(u = 1\sqrt{3} = \sqrt{3}\)
• Question 6: Given hypotenuse \(hyp = 6\sqrt{3}\).
• \(2x = 6\sqrt{3} \implies x = 3\sqrt{3}\) (short leg)
• Short leg \(y = 3\sqrt{3}\)
• Long leg \(x = (3\sqrt{3})\sqrt{3} = 9\)

Solve questions 7 through 9

• Question 7: Given long leg \(LL = 5\sqrt{3}\).
• \(x\sqrt{3} = 5\sqrt{3} \implies x = 5\) (short leg)
• Short leg \(y = 5\)
• Hypotenuse \(x = 2 \cdot 5 = 10\)
• Question 8: Given long leg \(LL = 10\sqrt{3}\).
• \(x\sqrt{3} = 10\sqrt{3} \implies x = 10\) (short leg)
• Short leg \(y = 10\)
• Hypotenuse \(x = 2 \cdot 10 = 20\)
• Question 9: Given long leg \(LL = 9\).
• \(x\sqrt{3} = 9 \implies x = \frac{9}{\sqrt{3}} = 3\sqrt{3}\) (short leg)
• Short leg \(y = 3\sqrt{3}\)
• Hypotenuse \(x = 2 \cdot 3\sqrt{3} = 6\sqrt{3}\)

Answer:

No.	Answer
2	\(x = 12\), \(y = 6\sqrt{3}\)
3	\(a = 2\sqrt{3}\), \(b = 3\)
4	\(a = 3\sqrt{3}\), \(b = 3\)
5	\(u = \sqrt{3}\), \(y = 1\)
6	\(x = 9\), \(y = 3\sqrt{3}\)
7	\(x = 10\), \(y = 5\)
8	\(x = 20\), \(y = 10\)
9	\(x = 6\sqrt{3}\), \(y = 3\sqrt{3}\)