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Question Number 89662 by 974342176 last updated on 18/Apr/20

Given that sin A=(1/2) and sin C=((√3)/2) without u  using calculator solve  a) tan (A+C)  b) cos(A−C)

$${Given}\:{that}\:\mathrm{sin}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{sin}\:{C}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{without}\:{u} \\ $$$${using}\:{calculator}\:{solve} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:\left({A}+{C}\right) \\ $$$$\left.{b}\right)\:{cos}\left({A}−{C}\right) \\ $$

Answered by TANMAY PANACEA. last updated on 18/Apr/20

sinA=(1/2)=sin30^o   →A=30^o   sinC=((√3)/2)=sin60^o  →C=60^o   tan(A+C)=tan(90^o )=∞  cos(A−C)=cos(−30^o )=cos(30^o )=((√3)/2)

$${sinA}=\frac{\mathrm{1}}{\mathrm{2}}={sin}\mathrm{30}^{{o}} \:\:\rightarrow{A}=\mathrm{30}^{{o}} \\ $$$${sinC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={sin}\mathrm{60}^{{o}} \:\rightarrow{C}=\mathrm{60}^{{o}} \\ $$$${tan}\left({A}+{C}\right)={tan}\left(\mathrm{90}^{{o}} \right)=\infty \\ $$$${cos}\left({A}−{C}\right)={cos}\left(−\mathrm{30}^{{o}} \right)={cos}\left(\mathrm{30}^{{o}} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by jagoll last updated on 19/Apr/20

how about C = 120^o    then tan (A+C) = tan 150^o    = −tan 30^o  = −(1/(√3))

$$\mathrm{how}\:\mathrm{about}\:\mathrm{C}\:=\:\mathrm{120}^{\mathrm{o}} \: \\ $$$$\mathrm{then}\:\mathrm{tan}\:\left(\mathrm{A}+\mathrm{C}\right)\:=\:\mathrm{tan}\:\mathrm{150}^{\mathrm{o}} \: \\ $$$$=\:−\mathrm{tan}\:\mathrm{30}^{\mathrm{o}} \:=\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$

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