KCET · Maths · Three Dimensional Geometry
In a triangle \( A B C, a[b \cos C-c \cos B]= \)
- A \( a^{2} \)
- B \( \mathrm{b}^{2} \)
- C \( 00 \)
- D \( b^{2}-c^{2} \)
Answer & Solution
Correct Answer
(D) \( b^{2}-c^{2} \)
Step-by-step Solution
Detailed explanation
Given that, \(a[b \cos C-c \cos B]\)
We know that, \(a=b \cos C+c \cos B\)
So, \((b \cos C+c \cos B)(b \cos C-c \cos B)\)
\(=b^{2} \cos ^{2} C-c^{2} \cos ^{2} B\)
\(=b^{2}\left(1-\sin ^{2} C\right)-c^{2}\left(1-\sin ^{2} B\right)\)
We know that, \(\sin B=\frac{b^{2}}{4 R^{2}}\) and \(\sin C=\frac{c^{2}}{4 R^{2}}\)
So, \(b^{2}\left(1-\frac{c^{2}}{4 R^{2}}\right)-c^{2}\left(1-\frac{b^{2}}{4 R^{2}}\right)\)
\(=b^{2}-\frac{b^{2} c^{2}}{4 R^{2}}-c^{2}+\frac{c^{2} b^{2}}{4 R^{2}}=b^{2}-c^{2}\)
We know that, \(a=b \cos C+c \cos B\)
So, \((b \cos C+c \cos B)(b \cos C-c \cos B)\)
\(=b^{2} \cos ^{2} C-c^{2} \cos ^{2} B\)
\(=b^{2}\left(1-\sin ^{2} C\right)-c^{2}\left(1-\sin ^{2} B\right)\)
We know that, \(\sin B=\frac{b^{2}}{4 R^{2}}\) and \(\sin C=\frac{c^{2}}{4 R^{2}}\)
So, \(b^{2}\left(1-\frac{c^{2}}{4 R^{2}}\right)-c^{2}\left(1-\frac{b^{2}}{4 R^{2}}\right)\)
\(=b^{2}-\frac{b^{2} c^{2}}{4 R^{2}}-c^{2}+\frac{c^{2} b^{2}}{4 R^{2}}=b^{2}-c^{2}\)
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