Question

Special Triangles

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent trigonometric ratios and the exact values

cos30° = $\frac{\sqrt{3} }{2}$ and tan30° = $\frac{1}{\sqrt{3} }$ , then

cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{6}{m}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

$\sqrt{3}$ × m = 12 ( divide both sides by $\sqrt{3}$ )

m = $\frac{12}{\sqrt{3} }$ ← rationalise by multiplying numerator/ denominator by $\sqrt{3}$ )

m = $\frac{12}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ = $\frac{12\sqrt{3} }{3}$ = 4$\sqrt{3}$

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tan30° = $\frac{opposite}{adjacent}$ = $\frac{n}{6}$ = $\frac{1}{\sqrt{3} }$ ( cross- multiply )

$\sqrt{3}$ × n = 6 ( divide both sides by $\sqrt{3}$ )

n = $\frac{6}{\sqrt{3} }$ ← rationalise the denominator

n = $\frac{6}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ = $\frac{6\sqrt{3} }{3}$ = 2$\sqrt{3}$

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