Question

In a triangle ABC with usual notations if $\angle A=30^{\circ}$, then the value of $(1+\frac{a}{c}+\frac{b}{c})(1+\frac{c}{b}-\frac{a}{b})=$

• A. $\sqrt{3}-2$
• B. $2+\sqrt{5}$
• C. $\sqrt{3}+2$
• D. $2-\sqrt{5}$

Hint:

$\frac{(b+c)^2 - a^2}{bc} = 2(1 + \cos A)$ is a standard identity in properties of triangles.

Answer 1

The Correct Option is C

Step 1: Concept
Use the Cosine Rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.

Step 2: Analysis
Simplify the expression: $(\frac{c+a+b}{c})(\frac{b+c-a}{b}) = \frac{(b+c+a)(b+c-a)}{bc}$
$= \frac{(b+c)^2 - a^2}{bc} = \frac{b^2 + c^2 + 2bc - a^2}{bc} = \frac{(b^2 + c^2 - a^2) + 2bc}{bc}$

Step 3: Calculation
From Cosine Rule: $b^2 + c^2 - a^2 = 2bc \cos A$.
Expression $= \frac{2bc \cos A + 2bc}{bc} = 2 \cos A + 2$.
Given $A = 30^{\circ}$, $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
Value $= 2(\frac{\sqrt{3}}{2}) + 2 = \sqrt{3} + 2$.

Step 4: Conclusion
Hence, the value is $\sqrt{3}+2$. Final Answer: (C)