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QveST [7]
3 years ago
8

If a scalene triangle has its measures 4 m, 11 m and 8 m, find the largest angle.

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

129.8 approximately

Step-by-step explanation:

So this sounds like a problem for the Law of Cosines. The largest angle is always opposite the largest side in a triangle.

So 11 is the largest side so the angle opposite to it is what we are trying to find. Let's call that angle, X.

My math is case sensitive.

X is the angle opposite to the side x.

Law of cosines formula is:

x^2=a^2+b^2-2ab \cos(X)

So we are looking for X.

We know x=11, a=4, and b=8 (it didn't matter if you called b=4 and a=8).

11^2=4^2+8^2-2(4)(8)\cos(X)

121=16+64-64\cos(X)

121=80-64\cos(X)

Subtract 80 on both sides:

121-80=-64\cos(X)

41=-64\cos(X)

Divide both sides by -64:

\frac{41}{-64}=\cos(X)

Now do the inverse of cosine of both sides or just arccos( )

[these are same thing]

\arccos(\frac{-41}{64})=X

Time for the calculator:

X=129.8 approximately

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Answer:

x ≈ 20.42, y ≈ 11.71

Step-by-step explanation:

Using the cosine ratio on the right triangle on the right, that is

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Using the sine ratio on the right triangle on the left, that is

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Multiply both sides by x

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x = \frac{11.71}{sin35} ≈ 20.42

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