If a=(b+c)cos⁡θ and sin⁡θ=kbcb+ccos⁡A2 then k=

Given a=(b+c)cos⁡θ⇒cos⁡θ=a(b+c)sin2⁡θ=1−cos2⁡θ=1−a2(b+c)2=(b+c)2−a2(b+c)2=(b+c+a)(b+c−a)(b+c)2=2s.(2s−2a)(b+c)2=4⋅s(s−a)bc⋅bc(b+c)2sin⁡θ=2bc(b+c)cos⁡A2

If a=(b+c)cos⁡θ and sin⁡θ=kbcb+ccos⁡A2 then k=

Given a=(b+c)cos⁡θ⇒cos⁡θ=a(b+c)sin2⁡θ=1−cos2⁡θ=1−a2(b+c)2=(b+c)2−a2(b+c)2=(b+c+a)(b+c−a)(b+c)2=2s.(2s−2a)(b+c)2=4⋅s(s−a)bc⋅bc(b+c)2sin⁡θ=2bc(b+c)cos⁡A2

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If $a = (b + c) \cos θ$ and $\sin θ = \frac{k \sqrt{b c}}{b + c} \cos \frac{A}{2}$ then k=

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Detailed Solution

Given $a = (b + c) \cos θ \Rightarrow \cos θ = \frac{a}{(b + c)}$

$\begin{array}{l} \sin^{2} θ = 1 - \cos^{2} θ = 1 - \frac{a^{2}}{(b + c)^{2}} \\ = \frac{(b + c)^{2} - a^{2}}{(b + c)^{2}} = \frac{(b + c + a) (b + c - a)}{(b + c)^{2}} \\ = \frac{2 s . (2 s - 2 a)}{(b + c)^{2}} = 4 \cdot \frac{s (s - a)}{b c} \cdot \frac{b c}{(b + c)^{2}} \\ \sin θ = \frac{2 \sqrt{b c}}{(b + c)} \cos \frac{A}{2} \end{array}$