Question

Fill up the table on finding complementary angle of given angle:

	$Angle-1$	$Angle-2$
(1)	${50}^{\circ }$	_________
(2)	____________	${63}^{\circ }$
(3)	${47}^{\circ }$	_________
(4)	${56}^{\circ }$	_________
(5)	__________	${12}^{\circ }$
(6)	${67}^{\circ }$	_________

Answer 1

We know that,

if two angles are complementary, then their sum is ${90}^o$.

$\left(1\right)$ Given, the angle is ${50}^o$.

Then, the complement of ${50}^0={90}^0-{50}^0={40}^0$.

Hence, the complement of ${50}^0$ is ${40}^0$.

$\left(2\right)$ Given, the angle is ${63}^o$.

Then, the complement of ${63}^0={90}^0-{63}^0={27}^0$.

Hence, the complement of ${63}^0$ is ${27}^0$.

$\left(3\right)$ Given, the angle is ${47}^o$.

Then, the complement of ${47}^0={90}^0-{47}^0={43}^0$.

Hence, the complement of ${47}^0$ is ${43}^0$.

$\left(4\right)$ Given, the angle is ${56}^o$.

Then, the complement of ${56}^0={90}^0-{56}^0={34}^0$.

Hence, the complement of ${56}^0$ is ${34}^0$.

$\left(5\right)$ Given, the angle is ${12}^o$.

Then, the complement of ${12}^0={90}^0-{12}^0={78}^0$.

Hence, the complement of ${12}^0$ is ${78}^0$.

$\left(6\right)$ Given, the angle is ${67}^o$.

Then, the complement of ${67}^0={90}^0-{67}^0={23}^0$.

Hence, the complement of ${67}^0$ is ${23}^0$.