What is the general solution for 15^2=12^2+20^2-2*12*20*cos(a) ?

The general solution for 15^2=12^2+20^2-2*12*20*cos(a) is a=0.84386…+2pin,a=2pi-0.84386…+2pin

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15^2=12^2+20^2-21220*cos(a)

1 5^{2} = 1 2^{2} + 2 0^{2} - 2 \cdot 12 \cdot 20 \cdot cos (a)

a = 0.84386 \dots + 2 πn, a = 2 π - 0.84386 \dots + 2 πn

Solution steps

1 5^{2} = 1 2^{2} + 2 0^{2} - 2 \cdot 12 \cdot 20 cos (a)

Switch sides

1 2^{2} + 2 0^{2} - 2 \cdot 12 \cdot 20 cos (a) = 1 5^{2}

Multiply the numbers:

2 \cdot 12 \cdot 20 = 480

1 2^{2} + 2 0^{2} - 480 cos (a) = 1 5^{2}

1 2^{2} = 144

144 + 2 0^{2} - 480 cos (a) = 1 5^{2}

2 0^{2} = 400

144 + 400 - 480 cos (a) = 1 5^{2}

Add the numbers:

144 + 400 = 544

- 480 cos (a) + 544 = 1 5^{2}

1 5^{2} = 225

- 480 cos (a) + 544 = 225

Move

544

to the right side

- 480 cos (a) = - 319

Divide both sides by

- 480

cos (a) = \frac{319}{480}

Apply trig inverse properties

a = arccos (\frac{319}{480}) + 2 πn, a = 2 π - arccos (\frac{319}{480}) + 2 πn

Show solutions in decimal form

a = 0.84386 \dots + 2 πn, a = 2 π - 0.84386 \dots + 2 πn

5 sec (θ) - 2 sec^{2} (θ) = tan^{2} (θ) - 1

cos (2 x) + cos (x) = 0, (0, 2 π)

cos (3 θ) = 1, 0 \leq θ \leq 2 π

3 \cdot sin (\frac{x}{2}) + cos (x) = 1

sin (2 x) = \frac{2}{3}

What is the general solution for 15^2=12^2+20^2-2*12*20*cos(a) ?
The general solution for 15^2=12^2+20^2-2*12*20*cos(a) is a=0.84386…+2pin,a=2pi-0.84386…+2pin