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Sagot :
Answer:
Solution given:
relationship between perpendicular and hypotenuse is given by sin angle
Sin 45=opposite/hypotenuse
[tex]\frac{\sqrt{2}}{2}=\frac{x}{\sqrt{5}}[/tex]
z=[tex]\frac{\sqrt{2}}{2}*\sqrt{5}[/tex]
x=[tex]\frac{\sqrt{10}}{2}[/tex]
The value of x is;[tex]\frac{\sqrt{10}}{2}[/tex]
Answer:
[tex]\frac{\sqrt{10} }{2}[/tex]
Step-by-step explanation:
on a 45 - 45 - 90 triangle if the sides are x then the hypotenuse is x[tex]\sqrt{2}[/tex]
so,
[tex]\sqrt{5}[/tex] = x[tex]\sqrt{2}[/tex] (Divide by [tex]\sqrt{2}[/tex] )
[tex]\frac{\sqrt{5} }{\sqrt{2} }[/tex] = x (rationalize the denominator)
[tex]\frac{\sqrt{5} }{\sqrt{2} }[/tex] · [tex]\frac{\sqrt{2} }{\sqrt{2} }[/tex] = [tex]\frac{\sqrt{10} }{\sqrt{4} }[/tex] = [tex]\frac{\sqrt{10} }{2}[/tex]
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