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andrew-mc [135]
3 years ago
15

Evaluate cos((Sin^-1)0)

Mathematics
1 answer:
Kamila [148]3 years ago
4 0
First evaluate
sin^{-1} (0)
what angles in domain of [-pi/2, pi/2] make sin = 0
\theta = 0
Now we can evaluate the cos function:
cos (0) = 1
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\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

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We know that ,

\bold{Perimeter \: of \: rectangle = 2(l + b)} \\

where <u>b </u><u>=</u><u> </u><u>width </u><u>/</u><u> </u><u>breadth</u> of rectangle

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<u>substituting</u><u> </u><u>the </u><u>values </u><u>in </u><u>the </u><u>formula </u><u>stated </u><u>above </u><u>,</u>

\bold{80  = 2(25 + b)} \\  \\\bold{ \implies \: 25 + b = \cancel \frac{80}{2} } \\  \\ \bold{\implies \: 25 + b = 40 }\\  \\ \bold{\implies \: b = 40 - 25 }\\  \\\bold{ \implies \: b = 15 \: feet}

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