In triangle $ABC$,if $a=13, b=8, c=7$,then $\cos(B+C) = $

In a triangle $ABC$,with usual notations,if $a=4, b=8, \angle C=60^{\circ}$,then the value of $\angle B$ and the ratio $\cos A : \cos C$ respectively are,

If in a triangle $\sin A : \sin C = \sin (A - B) : \sin (B - C)$,then $a^2, b^2, c^2$:

Let $ABC$ be a triangle with $\angle C=90^{\circ}$. Draw $CD$ perpendicular to $AB$. Choose points $M$ and $N$ on sides $AC$ and $BC$ respectively such that $DM$ is parallel to $BC$ and $DN$ is parallel to $AC$. If $DM=5$ and $DN=4$,then $AC$ and $BC$ are respectively equal to:

If in a triangle $ABC$,side $a = (\sqrt{3} + 1) \text{ cm}$,$\angle B = 30^\circ$,and $\angle C = 45^\circ$,then the area of the triangle is:

$ABC$ is a right-angled isosceles triangle with $\angle B = 90^\circ$. If $D$ is a point on $AB$ such that $\angle DCB = 15^\circ$ and $AD = 35 \, cm$,then $CD = $