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Fed [463]
3 years ago
5

If each two complementary angles has the same measure, then each angle will equal what?​

Mathematics
2 answers:
svetoff [14.1K]3 years ago
4 0
<h3>Answer: 45 degrees</h3>

=====================================================

Work Shown:

Let x be the measure of each of the two angles (which is valid because they both have the same measure). Since they are complementary, the angles add to 90

So,

x+x = 90

2x = 90

x = 90/2

x = 45

Each angle is 45 degrees

45+45 = 90

Pavlova-9 [17]3 years ago
3 0

Answer:

ok

Step-by-step explanation:

it's ammmmmmmmmmmmmmmmmmmmmost doen

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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS ASAP
mr_godi [17]

Answer:

x=46

Step-by-step explanation:

4(x+1)-3x=50

first you gotta take care of the () so multiple by 4

(4x+4) -3x=50

combine like terms, subtract 4-3 to get 1 so you have 1x left

4 + x = 50

and then solve for x

x=46

( i checked it to make sure it was right so dw)

6 0
2 years ago
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