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solve for x and round if necessary

Solve For X And Round If Necessary class=

Sagot :

DᴀʀᴋPᴀʀᴀᴅᴏxidn

[tex]\qquad\qquad\huge\underline{{\sf Answer}}☂[/tex]

Let's solve ~

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

[tex]\qquad \sf  \dashrightarrow \: \frac{3}{5} = \frac{0.6}{x} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 3x = 5 \times 0.6[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 3 \div 3[/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 1[/tex]

Therefore, the required value of x is 1

BasketballEinsteinidn

Answer:

[tex]\displaystyle 1[/tex]

Step-by-step explanation:

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

OR

[tex]\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[/tex]

Information on trigonometric ratios

[tex]\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[/tex]

I am joyous to assist you at any time.

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