Question

How do I complete these questions a bit confused . (extra credit work )​

Answer 1

Answer:

(1)

$a = \frac{3\sqrt 3}{2}$

$b = \frac{3}{2}$

(2)

$a = \sqrt 6$

$b = \sqrt 2$

Step-by-step explanation:

Solving (1):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{3}$

Solve for a

$a = 3 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 3 * \frac{\sqrt 3}{2}$

$a = \frac{3\sqrt 3}{2}$

To solve for b, we make use of Pythagoras theorem

$3^2 = a^2 + b^2$

This gives

$3^2 = (\frac{3\sqrt 3}{2})^2 + b^2$

$9 = \frac{9*3}{4} + b^2$

$9 = \frac{27}{4} + b^2$

Collect like terms

$b^2 = 9 - \frac{27}{4}$

Take LCM and solve

$b^2 = \frac{36 - 27}{4}$

$b^2 = \frac{9}{4}$

Take square roots

$b = \frac{3}{2}$

Solving (2):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{2\sqrt 2}$

Solve for a

$a = 2\sqrt 2 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 2\sqrt 2 * \frac{\sqrt 3}{2}$

$a = \sqrt 2 * \sqrt 3$

$a = \sqrt 6$

To solve for b, we make use of Pythagoras theorem

$(2\sqrt 2)^2 = a^2 + b^2$

This gives

$(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2$

$8 = 6 + b^2$

Collect like terms

$b^2 = 8 - 6$

$b^2 = 2$

Take square roots

$b = \sqrt 2$