Geometry Problem. Please help.

DC = 10

BD = x

By 45 45 90 triangle side ratios, AB = x too. 

By 30 60 90 triangle side ratios, BC/AB = sqrt(3). 

(x+10)/x = sqrt(3)

x + 10 = xsqrt(3)

10 = x(sqrt(3) - 1)

10/(sqrt(3) - 1) = x

(10sqrt(3) + 10)/2 = x

5sqrt(3) + 5 = x

 

=^._.^=

In triangle ABC, angle B = 90 degrees and angle C = 30 degrees. Point D is on line segment BC such that angle ADB = 45 degrees, and DC=10. What is AB?

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∠B = 90º        ∠C = 30º        ∠ADB = 45º        DC = 10

 

Let E be the point on BC so that DE ⊥ BC.

 

DE = 10(sin30)

 

EC = sqrt(10² - 5²)

 

Angle DBE = 15º

 

BE = DE / tan15º

 

AB / BC = DE / EC

 

AB = 15.77350269