TS EAMCET · Maths · Properties of Triangles
In \(\triangle A B C\), if \(B+C=72^{\circ}\), then \(\left(1+\frac{a}{c}+\frac{b}{c}\right)\) \(\left(1+\frac{c}{b}-\frac{a}{b}\right)\) is equal to
- A \(2+\sqrt{5}\)
- B \(\frac{\sqrt{5}+1}{2 \sqrt{2}}\)
- C \(\frac{\sqrt{5}-2}{4}\)
- D \(\frac{5-\sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{5-\sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=\left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right) \\ & =\frac{(b+c)^2-a^2}{b c}=\frac{\left(b^2+c^2-a^2\right)+2 b c}{b c}\end{aligned}\)…
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