Question:medium

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

Show Hint

$\frac{(b+c)^2 - a^2}{bc} = 2(1 + \cos A)$ is a standard identity in properties of triangles.
Updated On: Jun 19, 2026
  • $\sqrt{3}-2$
  • $2+\sqrt{5}$
  • $\sqrt{3}+2$
  • $2-\sqrt{5}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
The goal is to simplify a given expression involving side lengths of a triangle $ABC$ with a known angle $\angle A = 30^\circ$.

Step 2: Key Formula or Approach:

We use the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.

Step 3: Detailed Explanation:

The expression is: \[ E = (1 + \frac{a}{c} + \frac{b}{c})(1 + \frac{c}{b} - \frac{a}{b}) \] \[ E = (\frac{c + a + b}{c})(\frac{b + c - a}{b}) \] \[ E = \frac{(b + c + a)(b + c - a)}{bc} \] \[ E = \frac{(b + c)^2 - a^2}{bc} \] \[ E = \frac{b^2 + c^2 + 2bc - a^2}{bc} \] From the Cosine Rule, $b^2 + c^2 - a^2 = 2bc \cos A$.
Substituting this into the expression: \[ E = \frac{2bc \cos A + 2bc}{bc} = 2(\cos A + 1) \] Given $\angle A = 30^\circ$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
\[ E = 2(\frac{\sqrt{3}}{2} + 1) = \sqrt{3} + 2 \]

Step 4: Final Answer:

The value is $\sqrt{3} + 2$.
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