If ABCD is a cyclic quadrilateral with AB=6, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In $	riangle ABC$,by the law of cosines:$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos 	heta$$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos 	heta = 36 + 16 - 48 \cos \

Vedclass · Accepted Answer

$\frac{3}{13}$

Vedclass · Answer

(A) In $	riangle ABC$,by the law of cosines:$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos 	heta$$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos 	heta = 36 + 16 - 48 \cos 	heta = 52 - 48 \cos 	heta \quad (i)$In a cyclic quadrilateral,opposite angles are supplementary. Thus,$\angle ADC = 180^{\circ} - 	heta$.In $	riangle ADC$,by the law of cosines:$AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos(180^{\circ} - 	heta)$Since $\cos(180^{\circ} - 	heta) = -\cos 	heta$,we have:$AC^2 = 3^2 + 5^2 - 2(3)(5)(-\cos 	heta) = 9 + 25 + 30 \cos 	heta = 34 + 30 \cos 	heta \quad (ii)$Equating $(i)$ and $(ii)$:$52 - 48 \cos 	heta = 34 + 30 \cos 	heta$$52 - 34 = 30 \cos 	heta + 48 \cos 	heta$$18 = 78 \cos 	heta$$\cos 	heta = \frac{18}{78} = \frac{3}{13}$

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

Similar Questions

In a triangle $ABC$, $b^2 \sin 2C + c^2 \sin 2B =$

In a triangle $ABC$,if $b=7, c=4\sqrt{3}$ and $A=\frac{\pi}{6}$,then $a \sin B \sin C =$

With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

In a $\Delta ABC$,$a, c, A$ are given and $b_1, b_2$ are two values of the third side $b$ such that $b_2 = 2b_1$. Then $\sin A = $

In a triangle $ABC$,with usual notations $a=2, b=3, c=5$,then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module