Question

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

Answer 1

(1)

`a = (3\sqrt 3)/(2)`

`b = (3)/(2)`

(2)

`a = \sqrt 6`

`b = \sqrt 2`

Explanation:

Solving (1):

Considering

`\theta = 60^o`

We have:

`\sin(\theta) = (Opposite)/(Hypotenuse)`

This gives:

`\sin(60^o) = (a)/(3)`

Solve for a

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

`\sin(60^o) = (\sqrt 3)/(2)`

So:

`a = 3 * (\sqrt 3)/(2)`

`a = (3\sqrt 3)/(2)`

To solve for b, we make use of Pythagoras theorem

`3^2 = a^2 + b^2`

This gives

`3^2 = ((3\sqrt 3)/(2))^2 + b^2`

`9 = (9*3)/(4) + b^2`

`9 = (27)/(4) + b^2`

Collect like terms

`b^2 = 9 - (27)/(4)`

Take LCM and solve

`b^2 = (36 - 27)/(4)`

`b^2 = (9)/(4)`

Take square roots

`b = (3)/(2)`

Solving (2):

Considering

`\theta = 60^o`

We have:

`\sin(\theta) = (Opposite)/(Hypotenuse)`

This gives:

`\sin(60^o) = (a)/(2\sqrt 2)`

Solve for a

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

`\sin(60^o) = (\sqrt 3)/(2)`

So:

`a = 2\sqrt 2 * (\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`