Question

In a quadrilateral ABCD , 3Â =2B =3Č =6Ď express all the angles in radians and degrees ​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In quadrilateral ABCD,}$

$\mathsf{3\angle{A}=2\angle{B}=3\angle{C}=6\angle{D}}$

$\underline{\textsf{To find:}}$

$\textsf{All the angles of Quadrilateral ABCD}$

$\underline{\textsf{Solution:}}$

$\textsf{We know that,}$

$\textsf{The sum of all interior angles of quadriateral is}\;\mathsf{360^{\circ}}$

$\textsf{Consider,}$

$\mathsf{3\angle{A}=2\angle{B}=3\angle{C}=6\angle{D}=k(say)}$

$\texsf{Then,}$

$\mathsf{\angle{A}=\dfrac{k}{3},\angle{B}=\dfrac{k}{2},\angle{C}=\dfrac{k}{3},\angle{D}=\dfrac{k}{6}}$

$\textsf{But}\;\mathsf{\angle{A}+\angle{B}+\angle{C}+\angle{D}=360^{\circ}}$

$\mathsf{\dfrac{k}{3}+\dfrac{k}{2}+\dfrac{k}{3}+\dfrac{k}{6}=360^{\circ}}$

$\mathsf{\dfrac{2k+3k+2k+k}{6}=360^{\circ}}$

$\mathsf{\dfrac{8k}{6}=360^{\circ}}$

$\mathsf{\dfrac{k}{6}=45^{\circ}}$

$\implies\mathsf{k=270^{\circ}}$

$\mathsf{\angle{A}=\dfrac{270^{\circ}}{3}=90^{\circ}=\dfrac{\pi}{2}}$

$\mathsf{\angle{B}=\dfrac{270^{\circ}}{2}=135^{\circ}=\dfrac{3\pi}{4}}$

$\mathsf{\angle{C}=\dfrac{270^{\circ}}{3}=90^{\circ}=\dfrac{\pi}{2}}$

$\mathsf{\angle{D}=\dfrac{270^{\circ}}{6}=45^{\circ}=\dfrac{\pi}{4}}$