Question

In a triangle ABC, if \( a = 5 \), \( b = 3 \), and \( c = 7 \), then the ratio:

\[ \sqrt{\frac{\sin(A - B)}{\sin(A + B)}} \]

• A. \(\frac{4}{7}\)
• B. 16
• C. 36
• D. \(\frac{4}{5}\)

Hint:

When dealing with trigonometric functions in a triangle, ensure to use the Law of Sines and Cosines properly, and simplify the expressions systematically to avoid errors.

Answer 1

The Correct Option is A

We are given the sides \(a = 5\), \(b = 3\), and \(c = 7\) in triangle \(ABC\). Our goal is to find the value of \(\sqrt{\frac{\sin(A-B)}{\sin(A+B)}}\).

 Step 1: Apply the Law of Cosines to find angle \(C\). The Law of Cosines states: \[ c^2 = a^2 + b^2 - 2ab \cos C \] Substituting the given values: \[ 7^2 = 5^2 + 3^2 - 2 \times 5 \times 3 \times \cos C \] \[ 49 = 25 + 9 - 30 \cos C \] \[ 49 = 34 - 30 \cos C \] \[ 30 \cos C = -15 \] \[ \cos C = -\frac{1}{2} \] Thus, \(C = 120^\circ\) (since \(C\) is obtuse).

 Step 2: Use the Law of Sines to find angles \(A\) and \(B\). The Law of Sines states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Substituting \(c = 7\) and \(C = 120^\circ\), we get: \[ \frac{5}{\sin A} = \frac{3}{\sin B} = \frac{7}{\sin 120^\circ} \] Since \(\sin 120^\circ = \frac{\sqrt{3}}{2}\), we have: \[ \frac{5}{\sin A} = \frac{7}{\frac{\sqrt{3}}{2}} = \frac{14}{\sqrt{3}} \] \[ \sin A = \frac{5 \sqrt{3}}{14} \] Similarly: \[ \frac{3}{\sin B} = \frac{7}{\frac{\sqrt{3}}{2}} = \frac{14}{\sqrt{3}} \] \[ \sin B = \frac{3 \sqrt{3}}{14} \]

 Step 3: Find \(\sin(A+B)\) and \(\sin(A-B)\). First, \(\sin(A+B)\): \[ A + B = 180^\circ - C = 180^\circ - 120^\circ = 60^\circ \] \[ \sin(A+B) = \sin 60^\circ = \frac{\sqrt{3}}{2} \] Next, \(\sin(A-B)\) can be computed using the identity: \[ \sin(A-B) = \sin A \cos B - \cos A \sin B \] Using the values of \(\sin A\) and \(\sin B\) calculated earlier, we find: \[ \sin(A-B) = \frac{5\sqrt{3}}{14} \times \frac{1}{2} - \frac{\sqrt{3}}{2} \times \frac{3\sqrt{3}}{14} \] \[ \sin(A-B) = \frac{5\sqrt{3}}{28} - \frac{9}{28} = \frac{5\sqrt{3} - 9}{28} \] 

Step 4: Compute Now, compute the ratio: \[ \frac{\sin(A-B)}{\sin(A+B)} = \frac{\frac{5\sqrt{3} - 9}{28}}{\frac{\sqrt{3}}{2}} = \frac{5\sqrt{3} - 9}{28} \times \frac{2}{\sqrt{3}} = \frac{2(5\sqrt{3} - 9)}{28\sqrt{3}} = \frac{4}{7} \] Thus: \[ \sqrt{\frac{\sin(A-B)}{\sin(A+B)}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}} \] Thus, the correct answer is: \[ \boxed{\frac{4}{7}} \]