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Question Number 16269 by Tinkutara last updated on 20/Jun/17

2^(nd) part of Q. 16214: Prove that r_1 = s tan ((A/2)), r_2 = s tan ((B/2)), r_3 = s tan ((C/2)).

$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{part}\:\mathrm{of}\:\mathrm{Q}.\:\mathrm{16214}:\:\mathrm{Prove}\:\mathrm{that} \\ $$$${r}_{\mathrm{1}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{A}}{\mathrm{2}}\right),\:{r}_{\mathrm{2}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{B}}{\mathrm{2}}\right), \\ $$$${r}_{\mathrm{3}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{C}}{\mathrm{2}}\right). \\ $$

Answered by mrW1 last updated on 20/Jun/17

Commented by mrW1 last updated on 20/Jun/17

with usage of diagram above from mr. ajfour we have AE=AB+BD AF=AC+CD r_1 =AE×tan (A/2) r_1 =AF×tan (A/2) ⇒2r_1 =(AE+AF)×tan (A/2) ⇒2r_1 =(AB+BC+AC)×tan (A/2) ⇒2r_1 =2s×tan (A/2) ⇒tan (A/2)=(r_1 /s) similarly ⇒tan (B/2)=(r_2 /s) ⇒tan (C/2)=(r_3 /s)

$$\mathrm{with}\:\mathrm{usage}\:\mathrm{of}\:\mathrm{diagram}\:\mathrm{above}\:\mathrm{from} \\ $$$$\mathrm{mr}.\:\mathrm{ajfour}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{AE}=\mathrm{AB}+\mathrm{BD} \\ $$$$\mathrm{AF}=\mathrm{AC}+\mathrm{CD} \\ $$$$\mathrm{r}_{\mathrm{1}} =\mathrm{AE}×\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\mathrm{r}_{\mathrm{1}} =\mathrm{AF}×\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2r}_{\mathrm{1}} =\left(\mathrm{AE}+\mathrm{AF}\right)×\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2r}_{\mathrm{1}} =\left(\mathrm{AB}+\mathrm{BC}+\mathrm{AC}\right)×\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2r}_{\mathrm{1}} =\mathrm{2s}×\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{r}_{\mathrm{1}} }{\mathrm{s}} \\ $$$$\mathrm{similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}=\frac{\mathrm{r}_{\mathrm{2}} }{\mathrm{s}} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\mathrm{C}}{\mathrm{2}}=\frac{\mathrm{r}_{\mathrm{3}} }{\mathrm{s}} \\ $$

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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