Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 188219 by Rupesh123 last updated on 26/Feb/23

Answered by mr W last updated on 26/Feb/23

Commented by mr W last updated on 27/Feb/23

2β+α=(π/2) ⇒β=(π/4)−(α/2) length of rectangle=a height of rectangle=b a=2+4 cos (30°−α) b=2+4 sin (30°−α) 2((√3)+(1/(tan β)))sin α=1+(1/(tan β)) 2((√3)+((1+tan (α/2))/(1−tan (α/2))))×((2 tan (α/2))/(1−tan^2 (α/2)))=1+((1+tan (α/2))/(1−tan (α/2))) 2((√3)+((1+t)/(1−t)))×((2t)/(1−t^2 ))=1+((1+t)/(1−t)) (2(√3)−3)t^2 −2(1+(√3))t+1=0 t=tan (α/2)=((1+(√3)−(√7))/(2(√3)−3)) ⇒α=2 tan^(−1) (((1+(√3)−(√7))/(2(√3)−3)))≈21.067630 a≈5.951489 b≈2.621074 A≈15.599293

$$\mathrm{2}\beta+\alpha=\frac{\pi}{\mathrm{2}}\:\Rightarrow\beta=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$${length}\:{of}\:{rectangle}={a} \\ $$$${height}\:{of}\:{rectangle}={b} \\ $$$${a}=\mathrm{2}+\mathrm{4}\:\mathrm{cos}\:\left(\mathrm{30}°−\alpha\right) \\ $$$${b}=\mathrm{2}+\mathrm{4}\:\mathrm{sin}\:\left(\mathrm{30}°−\alpha\right) \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\right)\mathrm{sin}\:\alpha=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta} \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)×\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}}=\mathrm{1}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)×\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}} \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right){t}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}} \\ $$$$\Rightarrow\alpha=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\right)\approx\mathrm{21}.\mathrm{067630} \\ $$$${a}\approx\mathrm{5}.\mathrm{951489} \\ $$$${b}\approx\mathrm{2}.\mathrm{621074} \\ $$$${A}\approx\mathrm{15}.\mathrm{599293} \\ $$

Commented by Rupesh123 last updated on 26/Feb/23

Very detailed solution! Perfect Explanation!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com