In a $\triangle ABC$,$a=1$,$b=\sqrt{3}$ and $\angle C=\pi/6$. Then the measure of the third side $c=$

In a $\triangle ABC$,the altitudes from $B$ and $C$ to the opposite sides are not shorter than their respective opposite sides. Then,one of the angles of $\triangle ABC$ is $........^{\circ}$

With usual notations,in triangle $ABC$,$a=\sqrt{3}+1$,$b=\sqrt{3}-1$ and $m \angle C=60^{\circ}$,then $A-B=$ (in $^{\circ}$)

In a triangle $ABC$, if $a=3, b=4$ and $\sin A=\frac{3}{4}$, then $\angle CBA = (\text{in } ^{\circ})?$

In $\triangle ABC$,if $b \cos \theta = c - a$,(where $\theta$ is an acute angle),then $(c - a) \tan \theta =$

What is the value of $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$?