Question

If 2a°, 3a°,4a° and 6a° are the the angles of a quadrilateral, find them
Please show full process​

Answer 1

Answer:

Given:-

Angles of a Quadiratral are=2a,3a,4a,6a

To find:-

Measures of each angle

Solution

As we know that in a Quadiratral

${\boxed{\sf Sum\:of\:angles=360°}}$

Substitute the values

$\qquad\quad {:}\longmapsto\sf 2a+3a+4a+6a=360$

Simplify

$\qquad\quad {:}\longmapsto\sf 5a+10a=360$

$\qquad\quad {:}\longmapsto\sf 15a=360$

$\qquad\quad {:}\longmapsto\sf a=\cancel{{\dfrac {360}{15}}}$

$\qquad\quad {:}\longmapsto\sf a=24$

__________________________

$\qquad\quad {:}\longmapsto\sf 2a=2×24=48°$

$\qquad\quad {:}\longmapsto\sf 3a=3×24=72°$

$\qquad\quad {:}\longmapsto\sf 4a=4×24=96°$

$\qquad\quad {:}\longmapsto\sf 6a=6×24=144°$

$\therefore\sf The \:angles\:are\;48°,72°,96°\:and\:144°.$

Answer 2

Step-by-step explanation:

2a°+3a°+4a°+6a°=360°( sum of angles of a quadrilateral)

so 15a°=360°

=>a=360/15

=>a=24°

so 2a=24×2=48°,3a=3×24=72°,4a=4×24=96;6a=6×24= 144°