Question

If $ABCD$ is a cyclic quadrilateral with $AB=6,BC=4,CD=5,DA=3$ and $\angle ABC$ = $\theta$ then $cos\theta$ =

• A. $\frac{3}{13}$
• B. $\frac{18}{76}$
• C. $\frac{16}{78}$
• D. $\frac{78}{86}$

Answer 1

Answer: (A)

In $\Delta ABC$,
$A{C}^2=A{B}^2+B{C}^2-2(AB)(AC)\cos \theta [$ By Cosine rule $]$
$\Rightarrow A{C}^2=6^2+4^2-2(6)(4)\cos \theta$
$\Rightarrow A{C}^2=36+16-48\cos \theta$
$\Rightarrow A{C}^2=52-48\cos \theta \dots$(i)
Now, In $\Delta ADC$
$A{C}^2=A{D}^2+D{C}^2-2(AD)(DC)\cos \left(180^{\circ }-\theta \right)$
$=3^2+5^2+2(3)(5)\cos \theta [$ By Cosine rule $]$
$A{C}^2=9+25+30\cos \theta$
$A{C}^2=34+30\cos \theta \dots$(ii)
By Eqs. (i) and (ii), we get
$52-48\cos \theta =34+30\cos \theta$
$\Rightarrow 52-34=48\cos \theta +30\cos \theta$
$\Rightarrow 18=78\cos \theta$
$\Rightarrow \cos \theta =\frac{18}{78}=\frac{3}{13}$