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We know angle A=45, angle B=\(30\) and BC=\sqrt{6}. What is AB?

 Feb 14, 2022
 #1
avatar+854 
+8

I am assuming that we have a triangle ABC even though not posted in the question. If that is so, then the first step is to identify a few extra triangles we can draw in. 

I drew an altitude from point C down to AB, and called the point of intersection D. 

 

Angle A is 45 degrees, so triangle ADC is a right isosceles triangle with hypotenuse AC. 

Angle B is 30 degrees, so triangle BCD is a 30-60-90 triangle with hypotenuse BC.

 

If those are the cases, then DC is \({\sqrt{6}\over2}\). We also know that AD = DC, so AD = \({\sqrt{6}\over2}\). BD is \(\sqrt{3}\) x DC, so BD is \(3\sqrt{2}\over2\).

 

What we also do know is that AB = BD + AD = \({\sqrt{6}\over2} + {3\sqrt{2}\over2}\).

 

Thus, \(AB = {\sqrt{6} + 3\sqrt{2}\over2}\).

 

PRsmileyYAcoolP

 Feb 14, 2022
 #2
avatar+36435 
+1

Using sin law

 

(first note that angle C = 180 - 45-30 = 105 degrees)

 

sin A / sqrt 6 =  sin 105 / AB

 

AB = sqrt 6   sin 105 / sin 45

 Feb 14, 2022

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