What is the value of (15)/(tan(60)) ?

The value of (15)/(tan(60)) is 5sqrt(3)

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(15)/(tan(60))

\frac{15}{tan ( 6 0 ^{\circ} )}

53

Decimal Notation

8.66025 \dots

Solution steps

\frac{15}{tan ( 6 0 ^{\circ} )}

Use the following trivial identity:

tan (6 0^{\circ}) = 3

tan (6 0^{\circ})

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 3

= \frac{15}{3}

Simplify

\frac{15}{3} : 53

\frac{15}{3}

Factor

15 : 3 \cdot 5

Factor

15 = 3 \cdot 5

= \frac{3 \cdot 5}{3}

Cancel

\frac{3 \cdot 5}{3} : 3 \cdot 5

\frac{3 \cdot 5}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3 \cdot 5}{3 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= 5 \cdot 3^{- \frac{1}{2} + 1}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 5 \cdot 3^{\frac{1}{2}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= 53

= 3 \cdot 5

= 53

2 sin (3 0^{\circ}) sin (6 0^{\circ})

3 sin (2) - 3 (2) + 3

sin (- 85 5^{\circ})

cos (24 5^{\circ})

csc^{2} (3 0^{\circ}) + tan^{2} (4 5^{\circ})