Question

In a cyclic quadrilateral $ABCD;a,b,c,d$denote the length of the sides $AB,BC,CD,\mathrm{and}DA$ respectively.

Then $\cos A$ is equal to.

• A. $\begin{array}{l}\frac{a^2+d^2-b^2-c^2}{2\left(bc+ad\right)}\end{array}$
• B. $\begin{array}{l}\frac{a^2+d^2-b^2-c^2}{2\left(bc-ad\right)}\end{array}$
• C. $\begin{array}{l}\frac{a^2-d^2+b^2-2c^2}{\left(bc+ad\right)}\end{array}$
• D. none of these

Answer 1

The correct option is A

$\begin{array}{l}\frac{a^2+d^2-b^2-c^2}{2\left(bc+ad\right)}\end{array}$

Explanation for the correct answer:

To find the value of $\cos A$:

From the above figure,

Consider triangle $ABD$,

$B{D}^2=a^2+d^2-2ad\cos A...\left(1\right)\left[\mathrm{By}\mathrm{cosine}\mathrm{rule}\right]$

Similarly from triangle $BDC$,

$\begin{array}{rcl}B{D}^2& =& b^2+c^2-2bc\cos C\\ & \Rightarrow & B{D}^2=b^2+c^2-2bc\cos \left(180-A\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}\mathbf{}\mathbf{i}\mathbf{s}\mathbf{}\mathbf{c}\mathbf{y}\mathbf{c}\mathbf{l}\mathbf{i}\mathbf{c}\mathbf{,}\mathbf{}\mathbf{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{180}}^{\mathbf{0}}\mathbf{]}\\ & \Rightarrow & a^2+d^2-2ad\cos A=b^2+c^2+2bc\cos A\mathbf{[}\mathbf{\because }\mathbf{f}\mathbf{r}\mathbf{o}\mathbf{m}\mathbf{}\mathbf{1}\mathbf{]}\\ & \Rightarrow & \cos A=\frac{a^2+d^2-b^2-c^2}{2\left(bc+ad\right)}\end{array}$

Hence, the correct option is A.