AB=3, DC=5, ∠ CAD=$45^o$, AB ⊥ BC. Find the length of AC.
It's inefficient but start by defining angle BCA to be C. Then angle CDA = 135 - C. Call x = AC. Then by Law of SinesAB=3, DC=5, ∠ CAD=$45^o$, AB ⊥ BC. Find the length of AC.
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I gave two equations. The first is the Law of Sines in triangle ABC and the second is the Law of Sines in triangle ADC. Perhaps I should have stated that.Thank you for your response. But we still need one more equation since we have 2 variables x and C.
Are you actually looking for an answer to this problem? Or do you have an elegant answer to share?
If the latter, then we might move this thread to the https://mathhelpboards.com/forums/challenge-questions-and-puzzles.28/ subforum.
Btw, at this time I could provide an answer, but it's rather long winded and I'm still looking for a more elegant solution.