Question

what are the possible remaining angle measures in triangle abc with ( mangle a = 40^{circ}), ( ac = 50), and ( bc = 45)?
( mangle bapprox121^{circ}), ( mangle capprox19^{circ}) and ( mangle bapprox59^{circ}), ( mangle capprox81^{circ})
( mangle bapprox121^{circ}), ( mangle capprox19^{circ})
( mangle bapprox59^{circ}), ( mangle capprox81^{circ})
no triangle can be formed with these measures.

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{\sin A}{BC}=\frac{\sin B}{AC}\). Given \(A = 40^{\circ}\), \(AC = 50\), and \(BC=45\). Substitute the values into the formula: \(\frac{\sin40^{\circ}}{45}=\frac{\sin B}{50}\). Then \(\sin B=\frac{50\sin40^{\circ}}{45}\). Calculate \(\sin40^{\circ}\approx0.6428\), so \(\sin B=\frac{50\times0.6428}{45}\approx0.7142\).

Step2: Find the possible values of \(B\)

Since \(\sin B\approx0.7142\), \(B=\sin^{- 1}(0.7142)\approx45.6^{\circ}\) or \(B = 180^{\circ}-45.6^{\circ}=134.4^{\circ}\).
Case 1: If \(B\approx45.6^{\circ}\approx46^{\circ}\) (approximate to whole number), then \(C=180^{\circ}-(40^{\circ}+46^{\circ}) = 94^{\circ}\) (not in the options).
Case 2: If \(B\approx134.4^{\circ}\approx134^{\circ}\) (approximate to whole number), then \(C=180^{\circ}-(40^{\circ}+134^{\circ})=6^{\circ}\) (not in the options).
Another way: Using the formula \(c^{2}=a^{2}+b^{2}-2ab\cos C\) (where \(a = BC = 45\), \(b = AC=50\), \(A = 40^{\circ}\)). But from the first - step (Law of Sines approach for angle), when we check the sum of angles in the given options:
For option 1: \(40 + 121+19=180\), but using Law of Sines \(\frac{\sin40}{45}
eq\frac{\sin121}{50}\) (since \(\frac{\sin40}{45}\approx0.0143\), \(\frac{\sin121}{50}=\frac{\sin(180 - 59)}{50}=\frac{\sin59}{50}\approx\frac{0.8572}{50}=0.0171\)).
For option 2: \(40+59 + 81=180\), \(\frac{\sin40}{45}\approx0.0143\), \(\frac{\sin59}{50}\approx0.0171
eq0.0143\).

Answer:

No triangle can be formed with these measures.