AP EAMCET · Maths · Properties of Triangles
In \(\triangle A B C\), if \(b+c: c+a: a+b=7: 8: 9\), then the smallest angle (in radians) of that triangle is
- A \(\cos ^{-1}\left(\frac{4}{5}\right)\)
- B \(\frac{\pi}{3}\)
- C \(\cos ^{-1}\left(\frac{3}{5}\right)\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{4}{5}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & b+c=7 k, c+a=8 k, a+b=9 k \\ & \Rightarrow 2(a+b+c)=24 k \Rightarrow a+b+c=12 k \\ & a=5 k ; b=4 k, c=3 k \\ & \text { Smallest angle }=C \\ & \Rightarrow \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{25+16-9}{2 \times 5 \times 4}=\frac{32}{40}=\frac{4}{5} \\ &…
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