Question-227357

Question Number 227357 by mr W last updated on 18/Jan/26

Answered by A5T last updated on 18/Jan/26

Let ∥ be x , tangent be t and radius be r. x(2x)=t^2 ⇒t=x(√2) (√((2r)^2 −r^2 ))=t−r⇒t=r(1+(√3)) ⇒cos?=(x/(2r))=(((t(√2))/2)/((2t)/(1+(√3))))=(((√2)(1+(√3)))/4) ⇒?=15°

$$\mathrm{Let}\:\parallel\:\mathrm{be}\:\mathrm{x}\:,\:\mathrm{tangent}\:\mathrm{be}\:\mathrm{t}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{be}\:\mathrm{r}. \\ $$$$\mathrm{x}\left(\mathrm{2x}\right)=\mathrm{t}^{\mathrm{2}} \Rightarrow\mathrm{t}=\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{2r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\mathrm{t}−\mathrm{r}\Rightarrow\mathrm{t}=\mathrm{r}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{cos}?=\frac{\mathrm{x}}{\mathrm{2r}}=\frac{\frac{\mathrm{t}\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\mathrm{2t}}{\mathrm{1}+\sqrt{\mathrm{3}}}}=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\mathrm{4}} \\ $$$$\Rightarrow?=\mathrm{15}° \\ $$

Commented by A5T last updated on 18/Jan/26

Commented by A5T last updated on 18/Jan/26

C on circumference and AD diameter ⇒ ∠ACD = 90° ⇒ ∠BCD=90° AC=CD=x ⇒ △ACD≅ △BCD ⇒ BD=2r ⇒cos ∠BCD = (x/(2r))

$$\mathrm{C}\:\mathrm{on}\:\mathrm{circumference}\:\mathrm{and}\:\mathrm{AD}\:\mathrm{diameter}\: \\ $$$$\Rightarrow\:\angle\mathrm{ACD}\:=\:\mathrm{90}°\:\Rightarrow\:\angle\mathrm{BCD}=\mathrm{90}° \\ $$$$\mathrm{AC}=\mathrm{CD}=\mathrm{x}\:\Rightarrow\:\bigtriangleup\mathrm{ACD}\cong\:\bigtriangleup\mathrm{BCD}\:\Rightarrow\:\mathrm{BD}=\mathrm{2r} \\ $$$$\Rightarrow\mathrm{cos}\:\angle\mathrm{BCD}\:=\:\frac{\mathrm{x}}{\mathrm{2r}}\: \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply