What is the value of sqrt(15^2+40^2-2(15)(40))cos(60) ?

The value of sqrt(15^2+40^2-2(15)(40))cos(60) is 25/2

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sqrt(15^2+40^2-2(15)(40))cos(60)

1 5^{2} + 4 0^{2} - 2 (15) (40) cos (6 0^{\circ})

\frac{25}{2}

Decimal Notation

12.5

Solution steps

1 5^{2} + 4 0^{2} - 2 (15) (40) cos (6 0^{\circ})

= 1 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40 cos (6 0^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40 \frac{1}{2}

Simplify

1 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40 \frac{1}{2} : \frac{25}{2}

1 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40 \frac{1}{2}

1 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40 = 25

1 5^{2} + 4 0^{2} - 2 \cdot 15 \cdot 40

1 5^{2} = 225

= 225 + 4 0^{2} - 2 \cdot 15 \cdot 40

4 0^{2} = 1600

= 225 + 1600 - 2 \cdot 15 \cdot 40

Multiply the numbers:

2 \cdot 15 \cdot 40 = 1200

= 225 + 1600 - 1200

Add/Subtract the numbers:

225 + 1600 - 1200 = 625

= 625

Factor the number:

625 = 2 5^{2}

= 2 5^{2}

Apply radical rule:

2 5^{2} = 25

= 25

= 25 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 25}{2}

Multiply the numbers:

1 \cdot 25 = 25

= \frac{25}{2}

= \frac{25}{2}

cos (0.1) - \frac{199}{200}

\frac{2 tan ( 8 0 ^{\circ} ) + tan ( 2 0 ^{\circ} )}{cot ( 4 0 ^{\circ} )}

arccos (- \frac{7}{10})

400 sin (4 0^{\circ})

- ln (cos (\frac{π}{3}))

What is the value of sqrt(15^2+40^2-2(15)(40))cos(60) ?
The value of sqrt(15^2+40^2-2(15)(40))cos(60) is 25/2