Answer:
(1)
[tex]a = \frac{3\sqrt 3}{2}[/tex]
[tex]b = \frac{3}{2}[/tex]
(2)
[tex]a = \sqrt 6[/tex]
[tex]b = \sqrt 2[/tex]
Step-by-step explanation:
Solving (1):
Considering
[tex]\theta = 60^o[/tex]
We have:
[tex]\sin(\theta) = \frac{Opposite}{Hypotenuse}[/tex]
This gives:
[tex]\sin(60^o) = \frac{a}{3}[/tex]
Solve for a
[tex]a = 3 * \sin(60^o)[/tex]
[tex]\sin(60^o) = \frac{\sqrt 3}{2}[/tex]
So:
[tex]a = 3 * \frac{\sqrt 3}{2}[/tex]
[tex]a = \frac{3\sqrt 3}{2}[/tex]
To solve for b, we make use of Pythagoras theorem
[tex]3^2 = a^2 + b^2[/tex]
This gives
[tex]3^2 = (\frac{3\sqrt 3}{2})^2 + b^2[/tex]
[tex]9 = \frac{9*3}{4} + b^2[/tex]
[tex]9 = \frac{27}{4} + b^2[/tex]
Collect like terms
[tex]b^2 = 9 - \frac{27}{4}[/tex]
Take LCM and solve
[tex]b^2 = \frac{36 - 27}{4}[/tex]
[tex]b^2 = \frac{9}{4}[/tex]
Take square roots
[tex]b = \frac{3}{2}[/tex]
Solving (2):
Considering
[tex]\theta = 60^o[/tex]
We have:
[tex]\sin(\theta) = \frac{Opposite}{Hypotenuse}[/tex]
This gives:
[tex]\sin(60^o) = \frac{a}{2\sqrt 2}[/tex]
Solve for a
[tex]a = 2\sqrt 2 * \sin(60^o)[/tex]
[tex]\sin(60^o) = \frac{\sqrt 3}{2}[/tex]
So:
[tex]a = 2\sqrt 2 * \frac{\sqrt 3}{2}[/tex]
[tex]a = \sqrt 2 * \sqrt 3[/tex]
[tex]a = \sqrt 6[/tex]
To solve for b, we make use of Pythagoras theorem
[tex](2\sqrt 2)^2 = a^2 + b^2[/tex]
This gives
[tex](2\sqrt 2)^2 = (\sqrt 6)^2 + b^2[/tex]
[tex]8 = 6 + b^2[/tex]
Collect like terms
[tex]b^2 = 8 - 6[/tex]
[tex]b^2 = 2[/tex]
Take square roots
[tex]b = \sqrt 2[/tex]
