What is the general solution for 170=(160^2)/(32)sin(2θ) ?

The general solution for 170=(160^2)/(32)sin(2θ) is θ=(0.21413…)/2+pin,θ= pi/2-(0.21413…)/2+pin

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Popular >

170=(160^2)/(32)sin(2θ)

Solution

170 = \frac{16 0 ^{2}}{32} sin (2 θ)

Solution

θ = \frac{0.21413 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.21413 \dots}{2} + πn

Degrees

θ = 6.13444 \dots^{\circ} + 18 0^{\circ} n, θ = 83.86555 \dots^{\circ} + 18 0^{\circ} n

Solution steps

170 = \frac{16 0 ^{2}}{32} sin (2 θ)

Switch sides

\frac{16 0 ^{2}}{32} sin (2 θ) = 170

Simplify

\frac{16 0 ^{2}}{32} : 800

\frac{16 0 ^{2}}{32}

Factor

16 0^{2} : 2^{10} \cdot 5^{2}

Factor

160 = 2^{5} \cdot 5

= (2^{5} \cdot 5)^{2}

Apply exponent rule:

(ab)^{c} = a^{c} b^{c}

= 5^{2} (2^{5})^{2}

Simplify

(2^{5})^{2} : 2^{10}

(2^{5})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{5 \cdot 2}

Multiply the numbers:

5 \cdot 2 = 10

= 2^{10}

= 2^{10} \cdot 5^{2}

Factor

32 : 2^{5}

Factor

32 = 2^{5}

= \frac{2 ^{10} \cdot 5 ^{2}}{2 ^{5}}

Cancel

\frac{2 ^{10} \cdot 5 ^{2}}{2 ^{5}} : 2^{5} \cdot 5^{2}

\frac{2 ^{10} \cdot 5 ^{2}}{2 ^{5}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{10}}{2 ^{5}} = 2^{10 - 5}

= 5^{2} \cdot 2^{10 - 5}

Subtract the numbers:

10 - 5 = 5

= 2^{5} \cdot 5^{2}

= 2^{5} \cdot 5^{2}

2^{5} = 32

= 5^{2} \cdot 32

5^{2} = 25

= 32 \cdot 25

Multiply the numbers:

32 \cdot 25 = 800

= 800

800 sin (2 θ) = 170

Divide both sides by

800

800 sin (2 θ) = 170

Divide both sides by

800

\frac{800 sin ( 2 θ )}{800} = \frac{170}{800}

Simplify

sin (2 θ) = \frac{17}{80}

sin (2 θ) = \frac{17}{80}

Apply trig inverse properties

sin (2 θ) = \frac{17}{80}

General solutions for

sin (2 θ) = \frac{17}{80}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 θ = arcsin (\frac{17}{80}) + 2 πn, 2 θ = π - arcsin (\frac{17}{80}) + 2 πn

2 θ = arcsin (\frac{17}{80}) + 2 πn, 2 θ = π - arcsin (\frac{17}{80}) + 2 πn

Solve

2 θ = arcsin (\frac{17}{80}) + 2 πn : θ = \frac{arcsin ( \frac{17}{80} )}{2} + πn

2 θ = arcsin (\frac{17}{80}) + 2 πn

Divide both sides by

2

2 θ = arcsin (\frac{17}{80}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{17}{80} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arcsin ( \frac{17}{80} )}{2} + πn

θ = \frac{arcsin ( \frac{17}{80} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{17}{80}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{17}{80} )}{2} + πn

2 θ = π - arcsin (\frac{17}{80}) + 2 πn

Divide both sides by

2

2 θ = π - arcsin (\frac{17}{80}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{17}{80} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} - \frac{arcsin ( \frac{17}{80} )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( \frac{17}{80} )}{2} + πn

θ = \frac{arcsin ( \frac{17}{80} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{17}{80} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.21413 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.21413 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 170=(160^2)/(32)sin(2θ) ?
The general solution for 170=(160^2)/(32)sin(2θ) is θ=(0.21413…)/2+pin,θ= pi/2-(0.21413…)/2+pin