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Find x.



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1.5
3 \/3
2

Find X 3 15 3 3 2 class=

Sagot :

[tex] x = \sqrt{3} [/tex]

Step-by-step explanation:

Given:

[tex] \sin(ABC) = \frac{AC}{AB} [/tex]

[tex] = > \sin(60) = \frac{3}{AB} [/tex]

[tex] = >AB= 2 \sqrt{3} [/tex]

Now by Pythagoras theorem,

[tex]AC^{2} + BC^{2} = AB^{2} [/tex]

Substituting values,

[tex] {3}^{2} + {x}^{2} = (2 \sqrt{3})^{2} [/tex]

[tex] = > {x} = \sqrt{3} [/tex]

Hence, value of x is [tex] \sqrt{3} [/tex]

Answer:

x = [tex]\sqrt{3}[/tex]

Step-by-step explanation:

Using the tangent ratio in the right triangle and the exact value

tan60° = [tex]\sqrt{3}[/tex] , then

tan60° = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{3}{x}[/tex] = [tex]\sqrt{3}[/tex] ( multiply both sides by x )

3 = x × [tex]\sqrt{3}[/tex] ( divide both sides by [tex]\sqrt{3}[/tex] )

[tex]\frac{3}{\sqrt{3} }[/tex] = x , then rationalising the denominator

x = [tex]\frac{3}{\sqrt{3} }[/tex] × [tex]\frac{\sqrt{3} }{\sqrt{3} }[/tex] = [tex]\frac{3\sqrt{3} }{3}[/tex] = [tex]\sqrt{3}[/tex]

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