If $\cos^2 A + \cos^2 C = \sin^2 B$,then $\Delta ABC$ is

Let $ABC$ be a triangle in which $AB=BC$. Let $X$ be a point on $AB$ such that $AX:XB=AB:AX$. If $AC=AX$,then the measure of $\angle ABC$ equals (in $^{\circ}$)

In a $\Delta ABC$,the ratio of angles $A : B : C = 3 : 5 : 4$. Then $a + b + c \sqrt{2}$ is equal to

In any triangle $ABC$,$a \cdot \cos^2 \frac{A}{2} + b \cdot \cos^2 \frac{B}{2} + c \cdot \cos^2 \frac{C}{2} =$

In a $\triangle ABC$,$\angle C = 60^{\circ}$ and $\angle A = 75^{\circ}$. If $D$ is a point on $AC$ such that the area of $\triangle BAD$ is $\sqrt{3}$ times the area of $\triangle BCD$,then the measure of $\angle ABD$ is (in $^{\circ}$)

In any $\triangle ABC$,the simplified form of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is