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solve the right triangle (45-45-90)

Solve The Right Triangle 454590 class=

Sagot :

Lublanaidn

Answer:

[tex]HI=HJ=\sqrt{30}[/tex]units

[tex]\angle I=45^{\circ}[/tex]

Step-by-step explanation:

We are given that

[tex]\angle J=45^{\circ}[/tex]

[tex]JI=2\sqrt{15}[/tex]units

We know that

[tex]sin\theta=\frac{Perpendicular\;side}{hypotenuse}[/tex]

Using the formula

[tex]sin45=\frac{HI}{2\sqrt{15}}[/tex]

[tex]\frac{1}{\sqrt{2}}\times 2\sqrt{15}=HI[/tex]

Where [tex]sin45^{\circ}=\frac{1}{\sqrt{2}}[/tex]

[tex]\frac{2\sqrt{15}\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=HI[/tex]

By using rationalization

[tex]HI=\sqrt{30}[/tex] units

[tex]Cos45=\frac{HJ}{2\sqrt{15}}[/tex]

Using the formula

[tex]cos\theta=\frac{base}{hypotenuse}[/tex]

[tex]\frac{1}{\sqrt{2}}\times 2\sqrt{15}=HJ[/tex]

[tex]\frac{2\sqrt{15}\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=HJ[/tex]

[tex]HJ=\sqrt{30}[/tex] units

When two sides are equal then angle made by two equal sides are equal

Therefore,

[tex]\angle J=\angle I=45^{\circ}[/tex]

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