MHT CET · Maths · Trigonometric Ratios & Identities
In \(\triangle \mathrm{ABC}\), with usual notations, if \(\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}\), then \(m \angle C\) is equal to
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c} \)
\( \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{3(a+b+c)}{a+b+c} \)
\( \frac{a}{b+c}+1+\frac{b}{c+a}+1=3 \)
\( \Rightarrow \frac{a}{b+c}+\frac{b}{c+a}=1 \)
\( \Rightarrow a(c+a)+b(b+c)=(b+c)(c+a) \)
\( \Rightarrow a c+a^2+b^2+b c=b c+a b+c^2+a c \)
\( \Rightarrow a^2+b^2-c^2=a b ...(i)\)
\(\therefore\) By cosine Rule,
\(\cos C=\frac{a^2+b^2-c^2}{2 a b} \)
\( \cos C=\frac{a b}{2 a b} \)
\( \Rightarrow \cos C=\frac{1}{2} \)
\( \Rightarrow C=\frac{\pi}{3}\) ...[From (i)]
\( \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{3(a+b+c)}{a+b+c} \)
\( \frac{a}{b+c}+1+\frac{b}{c+a}+1=3 \)
\( \Rightarrow \frac{a}{b+c}+\frac{b}{c+a}=1 \)
\( \Rightarrow a(c+a)+b(b+c)=(b+c)(c+a) \)
\( \Rightarrow a c+a^2+b^2+b c=b c+a b+c^2+a c \)
\( \Rightarrow a^2+b^2-c^2=a b ...(i)\)
\(\therefore\) By cosine Rule,
\(\cos C=\frac{a^2+b^2-c^2}{2 a b} \)
\( \cos C=\frac{a b}{2 a b} \)
\( \Rightarrow \cos C=\frac{1}{2} \)
\( \Rightarrow C=\frac{\pi}{3}\) ...[From (i)]
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