What is the value of sqrt(13^2+9^2-2(13)(9)*cos(120)) ?

The value of sqrt(13^2+9^2-2(13)(9)*cos(120)) is square root of 367

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sqrt(13^2+9^2-2(13)(9)*cos(120))

1 3^{2} + 9^{2} - 2 (13) (9) \cdot cos (12 0^{\circ})

367

Decimal Notation

19.15724 \dots

Solution steps

1 3^{2} + 9^{2} - 2 (13) (9) cos (12 0^{\circ})

= 1 3^{2} + 9^{2} - 2 \cdot 13 \cdot 9 cos (12 0^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

= 1 3^{2} + 9^{2} - 2 \cdot 13 \cdot 9 (- \frac{1}{2})

Simplify

1 3^{2} + 9^{2} - 2 \cdot 13 \cdot 9 (- \frac{1}{2}) : 367

1 3^{2} + 9^{2} - 2 \cdot 13 \cdot 9 (- \frac{1}{2})

Apply rule

- (- a) = a

= 1 3^{2} + 9^{2} + 2 \cdot 13 \cdot 9 \cdot \frac{1}{2}

2 \cdot 13 \cdot 9 \cdot \frac{1}{2} = 117

2 \cdot 13 \cdot 9 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 \cdot 13 \cdot 9}{2}

Cancel the common factor:

2

= 1 \cdot 13 \cdot 9

Multiply the numbers:

1 \cdot 13 \cdot 9 = 117

= 117

= 1 3^{2} + 9^{2} + 117

1 3^{2} = 169

= 169 + 9^{2} + 117

9^{2} = 81

= 169 + 81 + 117

Add the numbers:

169 + 81 + 117 = 367

= 367

= 367

sin (22 5^{\circ}) cos (24 0^{\circ}) tan (12 0^{\circ})

3 cos (- 2)

\frac{800 \cdot 18 \cdot cos ( 1 5 ^{\circ} )}{7 \cdot cos ( 5 )}

sin (7 0^{\circ}) cos (1 5^{\circ}) + cos (7 0^{\circ}) sin (1 5^{\circ})

arccos (\frac{29}{40})

What is the value of sqrt(13^2+9^2-2(13)(9)*cos(120)) ?
The value of sqrt(13^2+9^2-2(13)(9)*cos(120)) is square root of 367