Answer:
[tex]a =22[/tex]
[tex]b = 11[/tex]
Step-by-step explanation:
Given
See attachment for triangle
Required
Find a and b
Using cosine formula, we have:
[tex]\cos \theta = \frac{Adjacent}{Hypotenuse}[/tex]
So, we have:
[tex]\cos (30) = \frac{11\sqrt 3}{a}[/tex]
Make a the subject
[tex]a = \frac{11\sqrt 3}{\cos (30)}[/tex]
[tex]\cos(30) = \frac{\sqrt 3}{2}[/tex]
So, we have:
[tex]a = \frac{11\sqrt 3}{\frac{\sqrt 3}{2}}[/tex]
Rewrite as:
[tex]a = 11\sqrt 3 \div \frac{\sqrt 3}{2}[/tex]
This gives:
[tex]a = 11\sqrt 3 * \frac{2}{\sqrt 3}[/tex]
[tex]a = 11 * 2[/tex]
[tex]a =22[/tex]
To solve for b, we use Pythagoras theorem
[tex]a^2 = b^2 + (11\sqrt 3)^2[/tex]
[tex]22^2 = b^2 + (11\sqrt 3)^2[/tex]
[tex]484 = b^2 + 363[/tex]
Collect like terms
[tex]b^2 = 484 - 363[/tex]
[tex]b^2 = 121[/tex]
Take positive square roots
[tex]b = \sqrt {121[/tex]
[tex]b = 11[/tex]
