Respuesta :

Hi there!  

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I believe your answer is:  

[tex]\text{D) } 40 \text{ and }50[/tex]

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Here’s why:  

  • Complementary Angles have a sum of 90°.

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[tex]\boxed{\text{Setting up an equation...}}}\\\\ \text{We know that complementary angles add up to 90 degrees.}\\\text{Therefore, the sum of the two angles is 90 degrees.}\\\\\rightarrow 4r + 5r = 90[/tex]

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[tex]\boxed{\text{Solving for 'r'...}}\\\\4r + 5r = 90\\---------------\\\rightarrow 9r = 90\\\\\rightarrow \frac{9r=90}{9}\\\\\boxed{r = 10}[/tex]

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[tex]\boxed{\text{Evaluating the expressions given...}}\\\\4r \text{ and } 5r\\----------------\\\boxed{4r}\\\\\rightarrow 4(10)\\\\\rightarrow 4 * 10\\\\\rightarrow\boxed{40}\\----------------\\\boxed{5r}\\\\\rightarrow 5(10)\\\\\rightarrow 5 * 10\\\\\rightarrow \boxed{50}[/tex]

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The correct answer should be D.

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Hope this helps you. I apologize if it’s incorrect.  

Answer:

D   40 degree and 50 degree

Step-by-step explanation:

*  We can work in units when dealing with tens as 4 + 5 = 9

* As the angle ASD is a Right Angle   4r  :  5r = 9r

* 2 angles we see within add up to 90 degree  90 = 9r

  ASM + MSO = 90 degree

* 5r = ASM   and 4r = MSO =  50 = ASM and 40 = MSO

 Therefore ASO - MSO = ASM   = 90 - 50 = 40

  90 - 50 = 40  and 90 - 5r = 4r

  4r = 40 degree and 5r = 50 degree

So just list a few of the above and show that

4r + 5r = 90

9r =  90/9  

r  = 10

Rounding to zero angles =  40 + 50 = 90 as r = 10

* Answer is D   40 degree and 50 degree

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