What is the value of sqrt(6^2+8^2+2*6*8*cos(90)) ?

The value of sqrt(6^2+8^2+2*6*8*cos(90)) is 10

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sqrt(6^2+8^2+268*cos(90))

6^{2} + 8^{2} + 2 \cdot 6 \cdot 8 \cdot cos (9 0^{\circ})

10

Solution steps

6^{2} + 8^{2} + 2 \cdot 6 \cdot 8 cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (9 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}

= 0

= 6^{2} + 8^{2} + 2 \cdot 6 \cdot 8 \cdot 0

Simplify

6^{2} + 8^{2} + 2 \cdot 6 \cdot 8 \cdot 0 : 10

6^{2} + 8^{2} + 2 \cdot 6 \cdot 8 \cdot 0

Apply rule

0 \cdot a = 0

= 6^{2} + 8^{2} + 0

6^{2} + 8^{2} + 0 = 6^{2} + 8^{2}

= 6^{2} + 8^{2}

6^{2} = 36

= 36 + 8^{2}

8^{2} = 64

= 36 + 64

Add the numbers:

36 + 64 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

= 10

tan (\frac{2 π}{3} + \frac{π}{4})

tan (0.24 π)

\frac{50}{tan ( 2 0 ^{\circ} )}

4 sec (4) (e^{4} + e^{4} tan (4))

cos (\frac{7 π}{6}) + i sin (\frac{7 π}{6})

What is the value of sqrt(6^2+8^2+2*6*8*cos(90)) ?
The value of sqrt(6^2+8^2+2*6*8*cos(90)) is 10