Question

determine the number of triangles abc possible with the given parts. a = 24, b = 20, a = 49° a. 2 b. 1 c. 3 d. 0

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Substitute \(a = 24\), \(b = 20\), \(A=49^{\circ}\). So \(\sin B=\frac{b\sin A}{a}\).

$$ \sin B=\frac{20\times\sin49^{\circ}}{24} $$

$$ \sin B=\frac{20\times0.7547}{24}\approx0.629 $$

Step2: Analyze the value of \(\sin B\)

Since \(0\lt\sin B = 0.629\lt1\), there are two possible values for \(B\) in the range \(0^{\circ}\lt B\lt180^{\circ}\). One is \(B_1=\sin^{- 1}(0.629)\approx39^{\circ}\) and the other is \(B_2 = 180^{\circ}-39^{\circ}=141^{\circ}\). Also, check the sum of angles. For \(B_1\), \(A + B_1=49^{\circ}+39^{\circ}=88^{\circ}\lt180^{\circ}\). For \(B_2\), \(A + B_2=49^{\circ}+141^{\circ}=190^{\circ}\gt180^{\circ}\), but wait, no! Wait, correction: \(A = 49^{\circ}\), \(a=24\gt b = 20\). When \(a\gt b\) and \(A\) is acute (\(A = 49^{\circ}\lt90^{\circ}\)), and \(\sin B=\frac{b\sin A}{a}\lt1\), there is one triangle. Because when \(a\gt b\), the larger side is opposite the larger angle. The formula \(\sin B=\frac{b\sin A}{a}\), and since \(a\gt b\), \(B\) must be acute.

Answer:

B. 1