MHT CET · Maths · Properties of Triangles
In a triangle \(\mathrm{ABC}\), with usual notations \(\angle A=60^{\circ}\), then \(\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=\)
- A \(\frac{3}{2}\)
- B \(\frac{1}{2}\)
- C 1
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=\frac{a+b+c}{c} \times \frac{b+c-a}{b}\) \(=\frac{(b+c)^2-a^2}{b c} \)
\( =\frac{b^2+c^2-a^2+2 b c}{b c}=2\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 \)
\( =2 \cos A+2=2 \cos 60^{\circ}+2=2 \times \frac{1}{2}+2=3\)
\( =\frac{b^2+c^2-a^2+2 b c}{b c}=2\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 \)
\( =2 \cos A+2=2 \cos 60^{\circ}+2=2 \times \frac{1}{2}+2=3\)
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