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Chapter 6: Problem 13

Solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. $$A=115^{\circ}, C=35^{\circ}, c=200$$

Short Answer

Expert verified
The solution to the triangle is \( A \approx 68^{\circ}, B = 50^{\circ}, C \approx 62^{\circ} \) and \( a=6, b \approx 4.6, c=5 \).

Step by step solution

01

Apply the Law of Sines

We start by applying the Law of Sines to find the angle 'A'. It is given by \( \frac{\sin A}{a} = \frac{\sin B}{c} \). This gives \( \sin A = \frac{a \sin B}{c} \). Substituting given values \( \sin A = \frac{6 \sin 50^{\circ}}{5} \) and calculating gives \( \sin A \approx 0.9314 \). Now find angle A by taking the arcsine of the calculated number. \( A = \text{arcsin}(0.9314) \), which gives \( A \approx 68^{\circ} \).
02

Find the third angle 'C'

Once we have determined angle 'A', we can find the third angle 'C' by using the fact that the sum of the angles in triangle equals 180掳. Hence, \( C = 180^{\circ} - (A + B) = 180^{\circ} - (68^{\circ} + 50^{\circ}) = 62^{\circ} \).
03

Apply the Law of Sines again to find side 'b'

Now, with known angles 'A' and 'C' and one known side 'a', we can now find side 'b' using the Law of Sines again. \( \frac{b}{\sin B} = \frac{a}{\sin A} \). Solving this for 'b' gives us \( b = a (\frac{\sin B}{\sin A})\). Substituting the given quantities and calculating gives \( b \approx 4.6 \).

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