Question

study guide practice for test 1
date:

1. solve for y.
2. be able to solve using the 30-60-90 special right triangle theorem.

formulas are: \\(hyp = 2 \cdot sl\\) and \\(ll = \sqrt{3} \cdot sl\\)
find a.

1. solve for x and y.

Explanation:

Identify the given values and target for Question 12

Using the 45-45-90 Triangle Theorem and Special Right Triangles knowledge points

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Identify the given values and target for Question 13

We are given a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) right triangle \(ABC\).

• The angle at \(A\) is \(60^\circ\), and the angle at \(C\) is \(30^\circ\).
• The side opposite the \(30^\circ\) angle is the shorter leg (\(sl\)), which is \(AB = a\).
• The side opposite the \(60^\circ\) angle is the longer leg (\(LL\)), which is \(BC = 4.8\) (or \(4\sqrt{3}\) depending on the print, but it reads as \(4.8\) or \(4\sqrt{3}\). Looking closely at the text, it is \(4\sqrt{3}\) written as "4 \sqrt{3}"). Let's use \(LL = 4\sqrt{3}\).
• We need to find the shorter leg \(a\).

Solve for a in Question 13

Using the formula for a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle:

$$ LL = \sqrt{3} \cdot sl $$

Substitute the known values:

$$ 4\sqrt{3} = \sqrt{3} \cdot a $$

Divide both sides by \(\sqrt{3}\):

$$ a = 4 $$

Identify the given values and target for Question 14

We are given a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) right triangle.

• The side opposite the \(30^\circ\) angle is the shorter leg (\(sl\)), which is labeled as \(7\).
• The hypotenuse is labeled as \(x\).
• The longer leg opposite the \(60^\circ\) angle is labeled as \(y\).

Solve for x and y in Question 14

Using the \(30^\circ\text{-}60^\circ\text{-}90^\circ\) formulas:

• For the hypotenuse \(x\):

$$ x = 2 \cdot sl = 2 \cdot 7 = 14 $$

• For the longer leg \(y\):

$$ y = \sqrt{3} \cdot sl = 7\sqrt{3} $$

Answer:

Question 12

\(y = 5\sqrt{2}\text{ cm}\)

Question 13

\(a = 4\)

Question 14

\(x = 14\), \(y = 7\sqrt{3}\)