Question

Two angles are complementary. One is 10° more than the other . Find the angles.​

Answer 1

Given: Two angles are complementary angles. & one angle is 10° more than the other angle.

Need to find: The angles ?

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

Let's say, that the two angles be x and y respectively.

Given that :

One angle is 10° more than the other angle.

Therefore :

$:\implies\sf\quad y = x + 10^\circ\qquad\qquad\qquad\qquad\sf\Bigg\lgroup\ eq^{n}\:(i)\Bigg\rgroup$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\underline{\bf{\dag} \:\frak{As\;we\;know\;that\: :}}$⠀

Two angles said to be complementary angles when they add up to 90°.

⠀

$:\implies\sf\quad First_{\:(angle)} + Second_{\:(angle)} = 90^\circ$

$:\implies\sf\quad x + y = 90^\circ$

$:\implies\sf\quad x + x + 10^\circ = 90^\circ\qquad\qquad\qquad\sf\Bigg\lgroup\ From\;eq^{n}\:(i)\Bigg\rgroup$

$:\implies\sf\quad 2x + 10^\circ = 90^\circ$

$:\implies\sf\quad 2x = 90^\circ - 10^\circ$

$:\implies\sf\quad 2x = 80^\circ$

$:\implies\sf\quad x = \cancel\dfrac{80^\circ}{2}$

$:\implies\quad\underline{\boxed{\pmb{\frak{x = 40^\circ}}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

⠀$\underline{\bf{\dag} \:\:\sf{Now,\;Calculating\;'y'\; Therefore\: :}}$⠀⠀⠀

$:\implies\sf\quad x + y = 90^\circ$

$:\implies\sf\quad 40^\circ + y = 90^\circ$

$:\implies\sf\quad y = 90^\circ - 40^\circ$

$:\implies\quad\underline{\boxed{\pmb{\frak{y = 50^\circ}}}}$

$\therefore{\underline{\sf{Hence,\;the\;angles\;are\;{\pmb{\sf{50^\circ}}} \;and\; {\pmb{\sf{40^\circ}}} \:respectively.}}}$

⠀⠀⠀⠀⠀

Answer 2

$\tt {\underline{ \underline{ \red{✠Answer:-}}}}$

Let the angles be x and (x+10)

By the problem:-

x+x+10=90

$\implies \tt{2x + 1 0= 90}$

$\implies \tt{2x = 90 - 10}$

$\implies \tt{x = \cancel \frac{80}{2} }$

$\implies \tt{x = 40}$

The first angle is:

x= 40°

The second angle is:

x+10 = (40+10)° = 50°

$\therefore \text{The angles are 40° and 50°}.$