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Question Number 10363 by Joel575 last updated on 05/Feb/17
ΔABC 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))
$$\Delta{ABC}\:\mathrm{with}\:{AB}\:=\:\mathrm{5},\:{BC}\:=\:\mathrm{7},\:{CA}\:=\:\mathrm{8} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\left(\mathrm{sin}\:{A}\:+\:\mathrm{sin}\:{B}\:+\:\mathrm{sin}\:{C}\right)\:.\:\left(\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right) \\ $$
Answered by mrW1 last updated on 06/Feb/17
a=BC=7 b=AC=8 c=AB=5 s=((a+b+c)/2)=((7+8+5)/2)=10 t=(√(s(s−a)(s−b)(s−c)))=(√(10×3×2×5))=10(√3) sin A=((2t)/(bc))=((2×10(√3))/(8×5))=((√3)/2) cos A=(√(1−(3/4)))=(1/2) sin B=((2t)/(ac))=((2×10(√3))/(7×5))=((4(√3))/7) cos B=(√(1−((48)/(49))))=(1/7) sin C=((2t)/(ba))=((2×10(√3))/(8×7))=((5(√3))/(14)) cos C=(√(1−((75)/(196))))=((11)/(14)) cot (A/2)=((1+cos A)/(sin A))=((1+(1/2))/((√3)/2))=(3/( (√3))) cot (B/2)=((1+cos B)/(sin B))=((1+(1/7))/((4(√3))/7))=(2/( (√3))) cot (C/2)=((1+cos C)/(sin C))=((1+((11)/(14)))/((5(√3))/(14)))=(5/( (√3))) (sin A+sin B+sin C)(cot (A/2)+cot (B/2)+cot (C/2)) =(((√3)/2)+((4(√3))/7)+((5(√3))/(14)))((3/( (√3)))+(2/( (√3)))+(5/( (√3)))) =((20(√3))/(14))×((10)/( (√3)))=((100)/7)
$${a}={BC}=\mathrm{7} \\ $$$${b}={AC}=\mathrm{8} \\ $$$${c}={AB}=\mathrm{5} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}=\frac{\mathrm{7}+\mathrm{8}+\mathrm{5}}{\mathrm{2}}=\mathrm{10} \\ $$$${t}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}=\sqrt{\mathrm{10}×\mathrm{3}×\mathrm{2}×\mathrm{5}}=\mathrm{10}\sqrt{\mathrm{3}} \\ $$$$\mathrm{sin}\:{A}=\frac{\mathrm{2}{t}}{{bc}}=\frac{\mathrm{2}×\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{8}×\mathrm{5}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:{A}=\sqrt{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:{B}=\frac{\mathrm{2}{t}}{{ac}}=\frac{\mathrm{2}×\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{7}×\mathrm{5}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:{B}=\sqrt{\mathrm{1}−\frac{\mathrm{48}}{\mathrm{49}}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\mathrm{sin}\:{C}=\frac{\mathrm{2}{t}}{{ba}}=\frac{\mathrm{2}×\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{8}×\mathrm{7}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}} \\ $$$$\mathrm{cos}\:{C}=\sqrt{\mathrm{1}−\frac{\mathrm{75}}{\mathrm{196}}}=\frac{\mathrm{11}}{\mathrm{14}} \\ $$$$\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{cos}\:{A}}{\mathrm{sin}\:{A}}=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cot}\:\frac{{B}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{cos}\:{B}}{\mathrm{sin}\:{B}}=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}}{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{7}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cot}\:\frac{{C}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{cos}\:{C}}{\mathrm{sin}\:{C}}=\frac{\mathrm{1}+\frac{\mathrm{11}}{\mathrm{14}}}{\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}}}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}\right)\left(\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right) \\ $$$$=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{7}}+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}}\right)\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{20}\sqrt{\mathrm{3}}}{\mathrm{14}}×\frac{\mathrm{10}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{100}}{\mathrm{7}} \\ $$
Commented by Joel575 last updated on 06/Feb/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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