Question

directions: find the value of each variable.
4.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
5.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
6.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
7.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
8.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
9.
$x = \underline{\quad\quad\quad}$
$y = \underline{\quad\quad\quad}$
© gina wilson (all things algebra®, llc), 2014-2016

Explanation:

Step1: Solve Q4 (30-60-90 triangle)

Adjacent to 60° is 5.
$x = 5\tan(60^\circ) = 5\sqrt{3}$
$y = \frac{5}{\cos(60^\circ)} = 10$

Step2: Solve Q5 (30-60-90 triangle)

Opposite 30° is 14.
$x = 2\times14 = 28$
$y = 14\tan(60^\circ) = 14\sqrt{3}$

Step3: Solve Q6 (30-60-90 triangle)

Hypotenuse is 32.
$x = 32\sin(30^\circ) = 16$
$y = 32\cos(30^\circ) = 16\sqrt{3}$

Step4: Solve Q7 (30-60-90 triangle)

Hypotenuse is 46.
$x = 46\sin(60^\circ) = 23\sqrt{3}$
$y = 46\cos(60^\circ) = 23$

Step5: Solve Q8 (30-60-90 triangle)

Opposite 60° is 20.
$y = \frac{20}{\tan(60^\circ)} = \frac{20\sqrt{3}}{3}$
$x = \frac{20}{\sin(60^\circ)} = \frac{40\sqrt{3}}{3}$

Step6: Solve Q9 (30-60-90 triangle)

Adjacent to 30° is $9\sqrt{3}$.
$x = 9\sqrt{3}\tan(30^\circ) = 9$
$y = \frac{9\sqrt{3}}{\cos(30^\circ)} = 18$

Answer:

1. $x = 5\sqrt{3}$, $y = 10$
2. $x = 28$, $y = 14\sqrt{3}$
3. $x = 16$, $y = 16\sqrt{3}$
4. $x = 23\sqrt{3}$, $y = 23$
5. $x = \frac{40\sqrt{3}}{3}$, $y = \frac{20\sqrt{3}}{3}$
6. $x = 9$, $y = 18$