Question

which of the following statements is/are TRUE?

Answer 1

ANSWER.

=> option [ B ] and [ C ] is correct answer.

EXPLANATION.

$\sf : \implies \: x \: , y \: and \: z \: be \: positive \: real \: number \\ \\ \sf : \implies \: assume \: x \: , \: y \: and \: z \: are \: the \: length \: of \: sides \: of \: triangle \\ \sf : \implies \: opposite \: to \: its \: angle \: x \: , \: y \: and \: z \:$

$\sf : \implies \: equation \: are \: = \tan( \dfrac{x}{z} ) + \tan( \dfrac{y}{z} ) = \dfrac{2y}{x + y + z}$

$\sf : \implies \: formula \: of \: \tan( \dfrac{a}{z} ) = \sqrt{ \dfrac{(s - b)(s - c)}{s(s - a)} } \\ \\ \\ \sf : \implies \: formula \: of \: \tan( \frac{c}{2} ) = \sqrt{ \frac{(s - a)(s - b)}{s(s - c)} } \\ \\ \sf : \implies \: s = \frac{a + b + c}{2}$

$\sf : \implies \: \sqrt{ \dfrac{(s - b)(s - c)}{s(s - a)} } \: \: + \: \: \sqrt{ \dfrac{(s - a)(s - b)}{s(s - c)} } = \dfrac{b}{s} \\ \\ \\ \sf : \implies \: \frac{ \sqrt{(s - b)} }{ \sqrt{s} } ( \frac{s - c + s - a}{ \sqrt{(s - a)(s - c)} } ) = \frac{b}{s} \\ \\ \sf : \implies \: 2s \: - (c + a) = b \\ \\ \sf : \implies \: \frac{ \sqrt{(s - b)} }{ \sqrt{s} } (\frac{b}{ \sqrt{(s - a)(s - c)} } ) = \frac{b}{s}$

$\sf : \implies \: \sqrt{s(s - b)} = \sqrt{(s - a)(s - c)} \\ \\ \sf : \implies \: {s}^{2} - bs \: = {s}^{2} - (a + c)s + ac \\ \\ \sf : \implies \: s(a + c - b) = ac \\ \\ \sf : \implies \: (a + c + b)(a +c - b) = 2ac \\ \\ \sf : \implies \: {a}^{2} + {c}^{2} = {b}^{2}$

$\sf : \implies \: it \: satisfied \: the \: option \: (b) \: \: and \: \: (c) \\ \\ \sf : \implies \: (b) = y = x + z \\ \\ \sf : \implies \: (c) = \tan( \frac{x}{z} ) = \frac{x}{y + z}$