Question

With usual notation, in triangle ABC , \(\mathrm{m} \angle \mathrm{A}=30^{\circ}\) then the value of \(\left(1+\frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}}\right)\left(1+\frac{\mathrm{c}}{\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{b}}\right)\) is equal to

• A. \(\frac{2+\sqrt{3}}{2}\)
• B. \(2+\sqrt{3}\)
• C. \(\frac{1+\sqrt{3}}{2}\)
• D. \(1+\sqrt{3}\)

Hint:

Combine each bracket into a single fraction and use the cosine rule.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Write the brackets as \(\dfrac{c+a+b}{c}\) and \(\dfrac{b+c-a}{b}\).

Step 2: Multiply
The product is \(\dfrac{(b+c)^{2}-a^{2}}{bc}=\dfrac{b^{2}+c^{2}-a^{2}+2bc}{bc}\).

Step 3: Use the Cosine Rule
\(b^{2}+c^{2}-a^{2}=2bc\cos A\), so the value is \(2\cos A+2\).

Step 4: Substitute
With \(A=30^{\circ}\): \(2\cdot\dfrac{\sqrt{3}}{2}+2=2+\sqrt{3}\), option (B).

Final Answer:
The expression equals \(2+\sqrt{3}\). \[ \boxed{\text{(B) }2+\sqrt{3}} \]