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taurus [48]
2 years ago
5

Solve for x and round if necessary

Mathematics
2 answers:
Nastasia [14]2 years ago
4 0

\qquad\qquad\huge\underline{{\sf Answer}}☂

Let's solve ~

\qquad \sf  \dashrightarrow \: \sin(37 \degree) =   \frac{0.6}{x}

\qquad \sf  \dashrightarrow \: \frac{3}{5}  =  \frac{0.6}{x}

\qquad \sf  \dashrightarrow \: 3x = 5 \times 0.6

\qquad \sf  \dashrightarrow \: x = 3 \div 3

\qquad \sf  \dashrightarrow \:x = 1

Therefore, the required value of x is 1

SashulF [63]2 years ago
3 0

Answer:

\displaystyle 1

Step-by-step explanation:

\displaystyle \frac{x}{0,6} = csc\:37 \hookrightarrow 0,6csc\:37 = x \hookrightarrow 0,9969840846... = x \\ \\ 1 ≈ x

<em>OR</em>

\displaystyle \frac{0,6}{x} = sin\:37 \hookrightarrow xsin\:37 = 0,6 \hookrightarrow \frac{0,6}{sin\:37} = x \hookrightarrow 0,9969840846... = x \\ \\ 1 ≈ x

Information on trigonometric ratios

\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ

I am joyous to assist you at any time.

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