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# Geometry

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

Feb 14, 2022

#1
+516
+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, $$AB = {\sqrt{6} + 3\sqrt{2}\over2}$$.

PRYAP

Feb 14, 2022
#2
+36417
+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