Question

In the diagram below, what are AC and BC?

Please don't round your answer (keep your answer as a square root) and solve with the Law of Sines/Cosines

Answer 1

BC14.5 AC 11.3

Explanation:

Answer 2

`\( BC \)` is
`\( (60)/(\sin 105^\circ) \)` and
`\( AC \)` is
`\( 30√(2) \)`.

To find the lengths of sides AC and BC in triangle ABC, we can use the Law of Sines.

The Law of Sines states:

`\[ (\sin A)/(a) = (\sin B)/(b) = (\sin C)/(c) \]`

In triangle ABC, given that
`\(AB = 30\)`,
`\( \angle BAC = 30^\circ\)`, and
`\( \angle BCA = 45^\circ\)`, we can find side AC using the Law of Sines:

`\[ (\sin \angle BAC)/(AB) = (\sin \angle BCA)/(AC) \]`

Substitute the given values:

`\[ (\sin 30^\circ)/(30) = (\sin 45^\circ)/(AC) \]`

Now, solve for AC:

`\[ AC = (30 \cdot \sin 45^\circ)/(\sin 30^\circ) \]`

`\[ AC = (30 \cdot (√(2))/(2))/((1)/(2)) \]`

`\[ AC = 30 \cdot √(2) \]`

So,
`\( AC = 30√(2) \)`.

Now, to find side BC, we can use the Law of Sines again:

`\[ (\sin \angle ABC)/(AC) = (\sin \angle BCA)/(BC) \]`

`\[ (\sin (180^\circ - \angle BAC - \angle BCA))/(AC) = (\sin 45^\circ)/(BC) \]`

`\[ (\sin 105^\circ)/(30√(2)) = (√(2))/(BC) \]`

Now, solve for BC:

`\[ BC = (30√(2) \cdot √(2))/(\sin 105^\circ) \]`

`\[ BC = (60)/(\sin 105^\circ) \]`

Using a calculator to find the exact value:

`\[ BC \approx (60)/(\sin 105^\circ) \]`

So,
`\( BC \)` is the reciprocal of the sine of
`\( 105^\circ \)`:

`\[ BC \approx (60)/(\sin 105^\circ) \]`

Therefore,
`\( BC \)` is
`\( (60)/(\sin 105^\circ) \)` and
`\( AC \)` is
`\( 30√(2) \)`.