ABC-with-AB-5-BC-7-CA-8-Find-the-value-of-sin-A-sin-B-sin-C-cot-A-2-cot-B-2-cot-C-2-

Question Number 10363 by Joel575 last updated on 05/Feb/17

Δ A B C with A B = 5, B C = 7, C A = 8

Find the value of

(\sin A + \sin B + \sin C) . (\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2})

Answered by mrW1 last updated on 06/Feb/17

a = B C = 7

b = A C = 8

c = A B = 5

s = \frac{a + b + c}{2} = \frac{7 + 8 + 5}{2} = 10

t = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{10 \times 3 \times 2 \times 5} = 10 \sqrt{3}

\sin A = \frac{2 t}{b c} = \frac{2 \times 10 \sqrt{3}}{8 \times 5} = \frac{\sqrt{3}}{2}

\cos A = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}

\sin B = \frac{2 t}{a c} = \frac{2 \times 10 \sqrt{3}}{7 \times 5} = \frac{4 \sqrt{3}}{7}

\cos B = \sqrt{1 - \frac{48}{49}} = \frac{1}{7}

\sin C = \frac{2 t}{b a} = \frac{2 \times 10 \sqrt{3}}{8 \times 7} = \frac{5 \sqrt{3}}{14}

\cos C = \sqrt{1 - \frac{75}{196}} = \frac{11}{14}

\cot \frac{A}{2} = \frac{1 + \cos A}{\sin A} = \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}}

\cot \frac{B}{2} = \frac{1 + \cos B}{\sin B} = \frac{1 + \frac{1}{7}}{\frac{4 \sqrt{3}}{7}} = \frac{2}{\sqrt{3}}

\cot \frac{C}{2} = \frac{1 + \cos C}{\sin C} = \frac{1 + \frac{11}{14}}{\frac{5 \sqrt{3}}{14}} = \frac{5}{\sqrt{3}}

(\sin A + \sin B + \sin C) (\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2})

= (\frac{\sqrt{3}}{2} + \frac{4 \sqrt{3}}{7} + \frac{5 \sqrt{3}}{14}) (\frac{3}{\sqrt{3}} + \frac{2}{\sqrt{3}} + \frac{5}{\sqrt{3}})

= \frac{20 \sqrt{3}}{14} \times \frac{10}{\sqrt{3}} = \frac{100}{7}

Commented by Joel575 last updated on 06/Feb/17

thank you sir

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