What is the general solution for solvefor θ,(90^2)/(250(32.2))=tan(θ) ?

The general solution for solvefor θ,(90^2)/(250(32.2))=tan(θ) is θ=0.78849…+pin

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solvefor θ,(90^2)/(250(32.2))=tan(θ)

solve for

θ, \frac{9 0 ^{2}}{250 ( 32.2 )} = tan (θ)

θ = 0.78849 \dots + πn

Degrees

θ = 45.17738 \dots^{\circ} + 18 0^{\circ} n

Solution steps

\frac{9 0 ^{2}}{250 ( 32.2 )} = tan (θ)

Switch sides

tan (θ) = \frac{9 0 ^{2}}{250 \cdot 32.2}

Simplify

\frac{9 0 ^{2}}{250 \cdot 32.2} : \frac{162}{161}

\frac{9 0 ^{2}}{250 \cdot 32.2}

Multiply the numbers:

250 \cdot 32.2 = 8050

= \frac{9 0 ^{2}}{8050}

Factor

9 0^{2} : 3^{4} \cdot 2^{2} \cdot 5^{2}

Factor

90 = 3^{2} \cdot 2 \cdot 5

= (3^{2} \cdot 2 \cdot 5)^{2}

Apply exponent rule:

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

= 2^{2} \cdot 5^{2} (3^{2})^{2}

Simplify

(3^{2})^{2} : 3^{4}

(3^{2})^{2}

Apply exponent rule:

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

= 3^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 3^{4}

= 3^{4} \cdot 2^{2} \cdot 5^{2}

Factor

8050 : 5^{2} \cdot 2 \cdot 7 \cdot 23

Factor

8050 = 5^{2} \cdot 2 \cdot 7 \cdot 23

= \frac{3 ^{4} \cdot 2 ^{2} \cdot 5 ^{2}}{5 ^{2} \cdot 2 \cdot 7 \cdot 23}

Cancel

\frac{2 ^{2} \cdot 3 ^{4} \cdot 5 ^{2}}{2 \cdot 5 ^{2} \cdot 7 \cdot 23} : \frac{3 ^{4} \cdot 2}{7 \cdot 23}

\frac{2 ^{2} \cdot 3 ^{4} \cdot 5 ^{2}}{2 \cdot 5 ^{2} \cdot 7 \cdot 23}

Cancel the common factor:

2

= \frac{3 ^{4} \cdot 5 ^{2} \cdot 2}{5 ^{2} \cdot 7 \cdot 23}

Cancel the common factor:

5^{2}

= \frac{3 ^{4} \cdot 2}{7 \cdot 23}

= \frac{3 ^{4} \cdot 2}{7 \cdot 23}

Multiply the numbers:

7 \cdot 23 = 161

= \frac{3 ^{4} \cdot 2}{161}

3^{4} \cdot 2 = 162

3^{4} \cdot 2

3^{4} = 81

= 81 \cdot 2

Multiply the numbers:

81 \cdot 2 = 162

= 162

= \frac{162}{161}

Apply trig inverse properties

tan (θ) = \frac{162}{161}

General solutions for

tan (θ) = \frac{162}{161}

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{162}{161}) + πn

θ = arctan (\frac{162}{161}) + πn

Show solutions in decimal form

θ = 0.78849 \dots + πn

sin (θ) = 2 sin (θ) cos (θ)

x, arccos (e^{- x}) = \frac{π}{3}

2 cos^{2} (4 x + 1) = 0

x, 2 sin (x) = - 1

t, f = a cos (\frac{t π}{6} - \frac{11 π}{12})

What is the general solution for solvefor θ,(90^2)/(250(32.2))=tan(θ) ?
The general solution for solvefor θ,(90^2)/(250(32.2))=tan(θ) is θ=0.78849…+pin