Question

In a quadrilateral
ABCD, 3A^=2B^=3c^=6D Express
all the angles in (1) degrees (2) radians​

Answer 1

$Given \: in \: a \: Quadrilateral \: ABCD \\3A = 2B = 3C = 6D$

$Let \: \blue { 3A = 2B = 3C = 6D = x }$

$i) 3A = x \implies A = \frac{x}{3} \: --(1)$

$ii) 2B = x \implies B = \frac{x}{2} \: --(2)$

$iii) 3C = x \implies C = \frac{x}{3} \: --(3)$

$ii) 6D = x \implies D = \frac{x}{6} \: --(4)$

$We \:know \: that ,$

$\pink{ \angle A + \angle B + \angle C + \angle D= 360\degree}$

$\implies \frac{x}{3} + \frac{x}{2} +\frac{x}{3} +\frac{x}{6} = 360$

$\implies \frac{2x+3x+2x+x}{6} = 360$

$\implies \frac{8x}{6} = 360$

$\implies x = 360 \times \frac{6}{8}$

$\implies x = 45 \times 6$

$\implies x = 270 \: --(5)$

$Now, 1. A = \frac{x}{3} \\=\frac{270}{3}\\= 90\degree$

$A = 90 \times \frac{\pi}{180} \\= \Big(\frac{\pi }{2}\Big)^{c}$

$2. B = \frac{x}{2} \\=\frac{270}{2}\\= 135\degree$

$A = 135 \times \frac{\pi}{180} \\= \Big(\frac{37\pi }{36}\Big)^{c}$

$3. C = \frac{x}{3} \\=\frac{270}{3}\\= 90\degree$

$C = 90\times \frac{\pi}{180} \\= \Big(\frac{\pi }{2}\Big)^{c}$

$4. D = \frac{x}{6} \\=\frac{270}{6}\\= 45\degree$

$D = 45 \times \frac{\pi}{180} \\= \Big(\frac{\pi }{4}\Big)^{c}$

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