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For the given right triangle, find the value x to the nearest tenth:​

For The Given Right Triangle Find The Value X To The Nearest Tenth class=

Sagot :

Hello!

For this type of problem, we are given a right triangle, and my go-to for finding a side length with another given side length and an angle value would most likely be the law of sines.

The law of sines states that:

[tex]\frac{a}{sin(A)}=\frac{b}{sin(B)}[/tex]

In the given triangle, the [tex]a[/tex] would be 45, and the opposite angle [tex]71[/tex], would be the [tex]A[/tex].

The same can be applied to the other side of the proportion.

[tex]x=b[/tex]

The opposite angle of side [tex]x[/tex] can be found using the definition of the combined angle of a triangle.

[tex]90+71+B=180[/tex]

[tex]B=19[/tex]

So now we can set up our proportion.

[tex]\frac{45}{sin(71)}=\frac{x}{sin(19)}[/tex]

[tex]x=sin(19)*\frac{45}{sin(71)}[/tex]

And we get that [tex]x[/tex] is around 15.5.

Hope this helps!

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