As shown in the figure, in the triangle ABC, the angle ACB is 90 degrees, the angle CD is high, and the angle a is 30 degrees  

As shown in the figure, in the triangle ABC, the angle ACB is 90 degrees, the angle CD is high, and the angle a is 30 degrees  

∵∠ACB=90°,∠A=30°
∴BC=1/2AB
∵ CD is high
∴CD⊥AB
∴∠CDB=90°
∴∠B+∠BCD=90°
∵∠B+∠A=90°
∴∠BCD=∠A=30°
∴BD=1/2BC
∵BC=1/2AB
∴BD=1/4AB

In triangle ABC, C = √ 3a, B = 30 °, then ∠ C is?

∵ - B = 30 °, from the sum theorem of triangle internal angles, there are: ∵ - a ＋ C = 150 °, ∵ - a = 150 ～ - C. ∵ C = √ 3a, ∵ from the sine theorem, it is easy to get: sin ∵ C = √ 3sin ∵ - A, ∵ sin ∵ C = √ 3sin (150 ～ - C), ∵ sin ∵ C = √ 3sin (30 ＋ C) = √ 3 (sin30 ° COSC)

In the triangle ABC, a = 27, C = 48, and C = 3A, find the length of B

From sine theorem
sinA/a=sinC/c
C=3A
sinA/27=sin3A/48
sin3A=3sinA-4(sinA)^3
Substitute:
27*（3sinA-4(sinA)^3）=48sinA
33sinA=108(sinA)^3
sinA≠0
The solution is: Sina = ± √ 11 / 6
Obviously, Sina > 0
Then Sina = √ 11 / 6
And a = 270
So cosa = √ (1 - (Sina) ^ 2) = 5 / 6
From the cosine theorem:
a^2=b^2+c^2-2bcccos∠A
Namely
b^2-80b+1575=0
The solution is b = 35, or 45