# What is the value of A (physics)?

Can someone please help for physics? I have an exam next week and my professor is not teaching the material:(

What is the value of A if sin(39.2)=cos(A)

Thanks:)

### 8 Answers

- 1 month agoFavorite Answer
sin(39.2) = cos(A)

sin(39.2) = sin(A + 90)

(39.2) = (A + 90)

A = 50.8

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- oldschoolLv 71 month ago
You can use a calculator to do this. Find the sine of 39.2° and then using the same calculator find the arcos of that. You will find that

A = 90°-39.2° = 50.8° since sin(Θ) = cos(90-Θ)

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- SpacemanLv 71 month ago
sin 39.2° = 0.632029303

cos A = 0.632029303

A = arccos 0.632029

A = 50.8 °

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- Anonymous1 month ago
the easier way :

sin 39.20° = 0.6320

0.6320 = cos A

A = arccos 0.6320 = 50.80°

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- skeptikLv 71 month ago
The most straightforward way of remembering things like this is to recall that these functions are all ratios involved in right triangles.

Sine = Opposite/Hypotenuse

Cosine = Adjacent/Hypotenuse.

Which means that for any given right triangle, where A and B are the two angles aside from the right angle, Sin(A) = Cos(B)

Also, since we know that the sum of all the internal angles of any triangle = 180 degrees, we know that the sum of A and B must be 90 degrees in a right triangle. In other words, they are complementary angles.

Combining all that previous knowledge together, we get Sin(A) = Cos(90-A). The Sine of any acute angle is the Cosine of its complementary angle.

Plugging your numbers into it, we get:

Sin(39.2) = Cos(90-39.2)

So the value of your answer is 90-39.2 = 50.8

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- LônLv 71 month ago
Easy.... it's 90 - 39.2 = 50.8.

sin (39.2) = 0.6320

cos (50.8) = 0.6320

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Sin A = Cos (90 - A)

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- billrussell42Lv 71 month ago
sin(39.2)=cos(A)

cosA = 0.6320293

A = 50.8º

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In this way unique inverses have been defined for inverses of the periodic functions.

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DId you miss the negative sign when subtracting?