If A, B, C, D are the angles of a cyclic quad | vedclass

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

Question

(D) Given that $ABCD$ is a cyclic quadrilateral.In a cyclic quadrilateral,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given that $ABCD$ is a cyclic quadrilateral.In a cyclic quadrilateral,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180^\circ$,we get $A = 180^\circ - C$.Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$.This implies $\cos A + \cos C = 0$.Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$.This implies $\cos B + \cos D = 0$.Adding these two equations,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

Similar Questions

$\sin {163^\circ} \cos {347^\circ} + \sin {73^\circ} \sin {167^\circ} = $

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

$\tan 75^\circ - \cot 75^\circ = $

If $\cot \theta + \cos \theta = p$ and $\cot \theta - \cos \theta = q$,then $(p^2 - q^2)^2$ in terms of $p$ and $q$ is: (in $pq$)

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \left( \frac{x + y}{2} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module