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Question Number 181741 by yaslm last updated on 29/Nov/22

Answered by cortano1 last updated on 30/Nov/22

 ⇒11 sin θ + 60 cos θ = 61 ; θ ∈(0,(π/2))  ⇒11 tan θ +60 = 61 sec θ   ⇒(11 tan θ+60)^2 = 3721 sec^2 θ  ⇒121 tan^2 θ+1320 tan θ+3600=3721 tan^2 θ+3721  ⇒3600 tan^2 θ−1320 tan θ+121=0  ⇒tan θ = ((11)/(60)) ⇒cot θ=((60)/(11))  ∴  (√(660 (tan θ+cot θ))) = (√(660(((11)/(60)) +((60)/(11)))))   = 61

$$\:\Rightarrow\mathrm{11}\:\mathrm{sin}\:\theta\:+\:\mathrm{60}\:\mathrm{cos}\:\theta\:=\:\mathrm{61}\:;\:\theta\:\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{11}\:\mathrm{tan}\:\theta\:+\mathrm{60}\:=\:\mathrm{61}\:\mathrm{sec}\:\theta\: \\ $$$$\Rightarrow\left(\mathrm{11}\:\mathrm{tan}\:\theta+\mathrm{60}\right)^{\mathrm{2}} =\:\mathrm{3721}\:\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\mathrm{121}\:\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{1320}\:\mathrm{tan}\:\theta+\mathrm{3600}=\mathrm{3721}\:\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{3721} \\ $$$$\Rightarrow\mathrm{3600}\:\mathrm{tan}\:^{\mathrm{2}} \theta−\mathrm{1320}\:\mathrm{tan}\:\theta+\mathrm{121}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta\:=\:\frac{\mathrm{11}}{\mathrm{60}}\:\Rightarrow\mathrm{cot}\:\theta=\frac{\mathrm{60}}{\mathrm{11}} \\ $$$$\therefore\:\:\sqrt{\mathrm{660}\:\left(\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\right)}\:=\:\sqrt{\mathrm{660}\left(\frac{\mathrm{11}}{\mathrm{60}}\:+\frac{\mathrm{60}}{\mathrm{11}}\right)} \\ $$$$\:=\:\mathrm{61}\: \\ $$

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