Chapter 13: Problem 7
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\)
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Chapter 13: Problem 7
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\)
These are the key concepts you need to understand to accurately answer the question.
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\((b-c) \cot \frac{A}{2}+(c-a) \cot \frac{B}{2}+(a-b) \cot \frac{C}{2}=0 .\)
The side of a base of a square pyramid is \(a\) meters and it's vertex is at a height of \(h\) meters above the center of the base. If \(\theta \& \phi\) be respectively the inclinations of any face to the base and of any two faces to one another, prove that \(\tan \theta=\frac{2 h}{a}\) and \(\cot \frac{\phi}{2}=\sqrt{1+\frac{a^{2}}{2 h^{2}}}\).
\(a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)
\(a^{2}=(b-c)^{2}+4 b c \sin ^{2} \frac{A}{2}\)
\(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\)
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