What is the general solution for 12^2=10^2+12^2-2(10)(12)cos(b) ?

The general solution for 12^2=10^2+12^2-2(10)(12)cos(b) is b=1.14102…+2pin,b=2pi-1.14102…+2pin

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12^2=10^2+12^2-2(10)(12)cos(b)

1 2^{2} = 1 0^{2} + 1 2^{2} - 2 (10) (12) cos (b)

b = 1.14102 \dots + 2 πn, b = 2 π - 1.14102 \dots + 2 πn

Solution steps

1 2^{2} = 1 0^{2} + 1 2^{2} - 2 (10) (12) cos (b)

Switch sides

1 0^{2} + 1 2^{2} - 2 \cdot 10 \cdot 12 cos (b) = 1 2^{2}

Subtract

1 2^{2}

from both sides

1 0^{2} + 1 2^{2} - 2 \cdot 10 \cdot 12 cos (b) - 1 2^{2} = 1 2^{2} - 1 2^{2}

Simplify

1 0^{2} - 2 \cdot 10 \cdot 12 cos (b) = 0

Multiply the numbers:

2 \cdot 10 \cdot 12 = 240

1 0^{2} - 240 cos (b) = 0

1 0^{2} = 100

100 - 240 cos (b) = 0

Move

100

to the right side

- 240 cos (b) = - 100

Divide both sides by

- 240

cos (b) = \frac{5}{12}

Apply trig inverse properties

b = arccos (\frac{5}{12}) + 2 πn, b = 2 π - arccos (\frac{5}{12}) + 2 πn

Show solutions in decimal form

b = 1.14102 \dots + 2 πn, b = 2 π - 1.14102 \dots + 2 πn

x, sin (t) - csc (x) + 1 = 0

tan^{2} (x) = sec^{2} (x) + 1

12 = 3 sec (θ) + 5 csc (θ)

tan^{2} (4 (x + \frac{π}{2})) = 3

sin (9 x + 2) = cos (6 x - 7)

What is the general solution for 12^2=10^2+12^2-2(10)(12)cos(b) ?
The general solution for 12^2=10^2+12^2-2(10)(12)cos(b) is b=1.14102…+2pin,b=2pi-1.14102…+2pin