What is the general solution for cos(C)=(5^2+4.05^2-3^2)/(2*(4.05*5)) ?

The general solution for cos(C)=(5^2+4.05^2-3^2)/(2*(4.05*5)) is C=0.64339…+2pin,C=2pi-0.64339…+2pin

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cos(C)=(5^2+4.05^2-3^2)/(2(4.055))

cos (C) = \frac{5 ^{2} + 4.0 5 ^{2} - 3 ^{2}}{2 \cdot ( 4.05 \cdot 5 )}

C = 0.64339 \dots + 2 πn, C = 2 π - 0.64339 \dots + 2 πn

Radians

C = 0.64339 \dots + 6.28318 \dots n, C = 5.63978 \dots + 6.28318 \dots n

Solution steps

cos (C) = \frac{5 ^{2} + 4.0 5 ^{2} - 3 ^{2}}{2 ( 4.05 \cdot 5 )}

Simplify

\frac{5 ^{2} + 4.0 5 ^{2} - 3 ^{2}}{2 \cdot 4.05 \cdot 5} : \frac{32.4025}{40.5}

\frac{5 ^{2} + 4.0 5 ^{2} - 3 ^{2}}{2 \cdot 4.05 \cdot 5}

5^{2} + 4.0 5^{2} - 3^{2} = 32.4025

5^{2} + 4.0 5^{2} - 3^{2}

5^{2} = 25

= 25 + 4.0 5^{2} - 3^{2}

4.0 5^{2} = 16.4025

= 25 + 16.4025 - 3^{2}

3^{2} = 9

= 25 + 16.4025 - 9

Add/Subtract the numbers:

25 + 16.4025 - 9 = 32.4025

= 32.4025

= \frac{32.4025}{2 \cdot 5 \cdot 4.05}

Multiply the numbers:

2 \cdot 4.05 \cdot 5 = 40.5

= \frac{32.4025}{40.5}

Apply trig inverse properties

cos (C) = \frac{32.4025}{40.5}

General solutions for

cos (C) = \frac{32.4025}{40.5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

C = arccos (\frac{32.4025}{40.5}) + 2 πn, C = 2 π - arccos (\frac{32.4025}{40.5}) + 2 πn

C = arccos (\frac{32.4025}{40.5}) + 2 πn, C = 2 π - arccos (\frac{32.4025}{40.5}) + 2 πn

Show solutions in decimal form

C = 0.64339 \dots + 2 πn, C = 2 π - 0.64339 \dots + 2 πn

- 1 + sin (x) = - 1.3

\frac{sin ( x )}{45} = \frac{sin ( 3 5 ^{\circ} )}{30}

cos (x) = \frac{4 ( - \frac{1}{2} ) - 1}{3}

1 = l sin (x), 2 = l cos (x)

y, t = 2 cot (6 0^{\circ} y + 3 0^{\circ})

What is the general solution for cos(C)=(5^2+4.05^2-3^2)/(2*(4.05*5)) ?
The general solution for cos(C)=(5^2+4.05^2-3^2)/(2*(4.05*5)) is C=0.64339…+2pin,C=2pi-0.64339…+2pin