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Given a = 54, b = 68, and c = 90, use the law of cosines

tofind the angleof the oblique triangle (not a righttriangle). Round to
the nearest whole number.

User JayV
by
4.0k points

1 Answer

3 votes

Answer:

94°

Explanation:

Oblique triangles are of two types: acute or obtuse triangle. They are not right triangles since they have no right angle.

This question can be answer using the Law of cosines, which states that:


\\ c^2 = a^2 + b^2 - 2a*b * cos(\gamma) [1]

Where
\\ \gamma is the angle of the respective opposite side c, a and b are the other triangles' sides.

Well, since we have all the information needed to find
\\ \gamma, that is, a = 54, b = 68 and c = 90, we calculate it using the formula [1] as follows:


\\ 90^2 = 54^2 + 68^2 - 2*(54)*(68)*cos(\gamma)

Solving this equation for
\\ cos(\gamma):


\\ 90^2 - 54^2 - 68^2 = - 2*(54)*(68)*cos(\gamma)


\\ 8100 - 2916 - 4624 = - 7344*cos(\gamma)


\\ 560 = - 7344*cos(\gamma)


\\ (560)/(-7344) = cos(\gamma)


\\ cos(\gamma) = - 0.07625 (for four significant figures)

To obtain
\\ \gamma, we need to determine which angle has a
\\ cos(\gamma) = -0.07625. For this, it is necessary the
\\ arccos(x) function, also expressed as
\\ cos^(-1)(x).

Then


\\ cos^(-1)[cos(\gamma)] = cos^(-1)(-0.07625)


\\ \gamma = cos^(-1)(-0.07625) = 94.37 degrees.

Rounding the result to the nearest whole number, the answer is
\\ \gamma = 94°.

The rest of the angles can be obtained using the Law of sines.

User Stuart Helwig
by
4.4k points