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Chapter 7: Problem 18

Solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. $$a=4, b=6, c=9$$

Short Answer

Expert verified
The measures for the angles in the triangle are \(A = 32藲\), \(B = 56藲\) and \(C = 92藲\).

Step by step solution

01

Solve for Angle A using The Law of Cosines

Applying the law of cosines: \(c^2 = a^2 + b^2 - 2*a*b*cos(C)\), which can be rearranged as \(cos(C) = (a^2 + b^2 - c^2) / (2*a*b)\). Substituting the given values, \(cos(C) = (4^2 + 6^2 - 9^2) / (2*4*6)\) which results in \(cos(C) = -0.046875\). Then, use an arccosine to find the degree value for C: \(C = arccos(-0.046875) = 92藲\) (rounded to the nearest degree).
02

Calculate Angle B using The Law of Sines

According to the law of sines, \(a/sin(A) = b/sin(B) = c/sin(C)\). Arranging for \(B\): \(sin(B) = b*sin(C)/a\). Substituting the given values and the value for C that was previously calculated, \(sin(B) = 6*sin(92) / 4 = 1.45\). Then, use arcsine to get the degree value for B: \(B = arcsin(1.45) = 56藲\) (rounded to the nearest degree).
03

Deduct Angle A

Since in any triangle, the sum of the angles equals 180 degrees, therefore, \(A = 180 - C - B = 180 - 92 - 56 = 32藲\).

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