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Question Number 69192 by TawaTawa last updated on 21/Sep/19
Answered by mr W last updated on 21/Sep/19
let ∠BAD=α side length =a (a/(sin (60+α)))=(3/(sin α)) ...(i) (a/(sin (60+30−α)))=(a/(cos α))=(5/(sin (30−α))) ...(ii) (i)/(ii): ⇒((cos α)/(sin (60+α)))=((3 sin (30−α))/(5 sin α)) ((cos α)/(((√3)/2)cos α+(1/2)sin α))=((3((1/2) cos α−((√3)/2)sin α))/(5 sin α)) ((4 cos α)/( (√3)cos α+sin α))=((3(cos α−(√3)sin α))/(5 sin α)) (4/( (√3)+tan α))=((3(1−(√3) tan α))/(5 tan α)) let t=tan α (4/( (√3)+t))=((3−3(√3) t)/(5t)) 3(√3)t^2 +26t−3(√3)=0 ⇒t=tan α=((√3)/9)⇒α=tan^(−1) ((√3)/9)≈10.9° from (i): a=((3 sin (60+α))/(sin α))=3(((√3)/(2 tan α))+(1/2))=15 a=3+x+5 ⇒x=a−8=7
$${let}\:\angle{BAD}=\alpha \\ $$$${side}\:{length}\:={a} \\ $$$$\frac{{a}}{\mathrm{sin}\:\left(\mathrm{60}+\alpha\right)}=\frac{\mathrm{3}}{\mathrm{sin}\:\alpha}\:\:\:\:…\left({i}\right) \\ $$$$\frac{{a}}{\mathrm{sin}\:\left(\mathrm{60}+\mathrm{30}−\alpha\right)}=\frac{{a}}{\mathrm{cos}\:\alpha}=\frac{\mathrm{5}}{\mathrm{sin}\:\left(\mathrm{30}−\alpha\right)}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\left(\mathrm{60}+\alpha\right)}=\frac{\mathrm{3}\:\mathrm{sin}\:\left(\mathrm{30}−\alpha\right)}{\mathrm{5}\:\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{cos}\:\alpha}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha}=\frac{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\alpha−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\alpha\right)}{\mathrm{5}\:\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{4}\:\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha}=\frac{\mathrm{3}\left(\mathrm{cos}\:\alpha−\sqrt{\mathrm{3}}\mathrm{sin}\:\alpha\right)}{\mathrm{5}\:\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha}=\frac{\mathrm{3}\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\alpha\right)}{\mathrm{5}\:\mathrm{tan}\:\alpha} \\ $$$${let}\:{t}=\mathrm{tan}\:\alpha \\ $$$$\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}+\mathrm{t}}=\frac{\mathrm{3}−\mathrm{3}\sqrt{\mathrm{3}}\:{t}}{\mathrm{5}{t}} \\ $$$$\mathrm{3}\sqrt{\mathrm{3}}{t}^{\mathrm{2}} +\mathrm{26}{t}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\approx\mathrm{10}.\mathrm{9}° \\ $$$${from}\:\left({i}\right): \\ $$$${a}=\frac{\mathrm{3}\:\mathrm{sin}\:\left(\mathrm{60}+\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:\mathrm{tan}\:\alpha}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{15} \\ $$$${a}=\mathrm{3}+{x}+\mathrm{5} \\ $$$$\Rightarrow{x}={a}−\mathrm{8}=\mathrm{7} \\ $$
Commented by TawaTawa last updated on 21/Sep/19
God bless you sir, i appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

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