Question

HOLY FAITH MATHEWS 80 28. 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?

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

Given that,

• One angle is 10° more than the other angle. Therefore:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Answer 2

Let,

• The one angle = a

• The complement of a = 90° - a

From the question,

$\sf\implies \: \: {a + (a-10\degree) = 90 \degree}$

$\sf\implies \: \: {2a-10\degree = 90 \degree}$

$\sf\implies \: \: {2a = 90\degree + 10\degree}$

$\sf\implies \: \: {a = \cancel\frac{100\degree}{2}}$

$\sf\large\pink{\implies\: \: \frak{a=50\degree}}$

Therefore,

• One angle = a = 50°

• Other angle = a - 10° = 50° - 10° = 40°