Half-Angle Formula and the Area of a Triangle

If $\mathrm{h}$ is the perpendicular distance from $\mathrm{A}$ on $\mathrm{BC}$ of a triangle $\mathrm{ABC}$, prove that $h=\frac{a\mathrm{sinBsin}C}{\sin (\mathrm{B}+\mathrm{C})}$.

If the sides of a triangle $\mathrm{ABC}$ are in A.P., then show that cot $\frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2}$ are also in A.P.

[x] $(b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b)\cot \frac{C}{2}=0$.

[ix]$\frac{b-c}{b}{\cos }^2\frac{\mathrm{A}}{2}+\frac{c-a}{b}{\cos }^2\frac{\mathrm{B}}{2}+\frac{a-b}{c}{\cos }^2\frac{\mathrm{C}}{2}=0$.

In a $△ABC, \tan \frac{A}{2}=\frac{5}{6}, \tan \frac{B}{2}=\frac{20}{37}$; prove that, $a+c=2b$.

In a right-angled triangle $ABC$, right angled at $A$, let $l$ be the incentre, If $IB=\sqrt{10}$ units and $IC=\sqrt{5}$ units, find the area of $△ABC$.

If $x, y$ and $z$ be the lengths of the perpendiculars from circum centre of a triangle $ABC$ on the sides $\overline{BC},\overline{ CA}$ and $\overline{AB}$ respectively, then show that
$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{abc}{4xyz}$.

If in a $△ABC$, two sides are $a=6, b=3$, and $\cos (A-B)=\frac{4}{5}$, then

If in a $∆ABC$, two sides are $a=3,b=5$, and $\cos (A-B)=\frac{7}{25}$. Then, find the value of $k$ if length of the side $c$ is $\sqrt{k}$

If in a $∆ABC$, two sides are $a=3,b=5$, and $\cos (A-B)=\frac{7}{25}$. Find $\tan \frac{C}{2}$.

If $∆ABC$ is right-angled at $C$. Prove that : $\tan \frac{A}{2}=\sqrt{\frac{c-b}{c+b}}$.

In a $∆ABC$, the tangent of half the difference of two angles is one-third the tangent of half the sum of the angles. Determine the ratio of the sides opposite to the angles.

Show that in $∆ABC$, $b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=s$.

Show that in $∆ABC$ $\left(\tan \frac{A}{2}\right)\left(\tan \frac{B}{2}\right)=\frac{a+b-c}{a+b+c}$.

With the usual notations prove that area of $∆ABC=s\left(s-a\right)\tan \frac{A}{2}=s\left(s-b\right)\tan \frac{B}{2}=s\left(s-c\right)\tan \frac{C}{2}$.

If in $∆ABC$, $a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=\frac{3b}{2}$, then prove that $a, b, c$ are in $\mathrm{A}.\mathrm{P}.$

Show that $\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{{∆}^2}{abcs}$ for$∆ABC$ , where $∆$ is the area of triangle.

In $∆ABC$, if $a=18, b=24$ and $c=30$, then find the value of $\tan A$.

In $∆ABC$, if $a=18, b=24$ and $c=30$, then find the value of $\sin A$.

In $∆ABC$, if $a=18, b=24$ and $c=30$, then find the value of $∆,$where $∆$ is area of triangle.       [Enter the value excluding units]