Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 29249 by ajfour last updated on 05/Feb/18

Commented by ajfour last updated on 05/Feb/18

$$Findsidelengths\mathit{a},\mathit{b},\mathit{c}of△ABC$$$$intermsof{\mathit{r}}_1,{\mathit{r}}_2,and{\mathit{r}}_3.$$$$Wemaycall{r}_{1}as\mathit{p},$$$$r_{2}as\mathit{q}and{r}_{3}as\mathit{r}forthesakeof$$$$convinience.$$

Answered by mrW2 last updated on 05/Feb/18

Commented by mrW2 last updated on 05/Feb/18

$$DE=r+p$$$$EF=p+q$$$$FD=q+r$$$$\alpha ={\sin }^{-1}\left(\frac{p-q}{p+q}\right)$$$$\beta ={\sin }^{-1}\left(\frac{q-r}{q+r}\right)$$$$\gamma ={\sin }^{-1}\left(\frac{r-p}{r+p}\right)$$$$$$$$In\mathrm{\Delta }DEFwehave$$$$\cos \mathrm{\angle }DEF=\frac{{(p+q)}^2+{(r+p)}^2-{(q+r)}^2}{2(p+q)(r+p)}$$$$\Rightarrow \cos \mathrm{\angle }DEF=\frac{p(p+q)+r(p-q)}{(p+q)(r+p)}$$$$\Rightarrow u=\mathrm{\angle }DEF={\cos }^{-1}\left[\frac{p(p+q)+r(p-q)}{(p+q)(r+p)}\right]$$$$similarly:$$$$\Rightarrow \cos \mathrm{\angle }EFD=\frac{q(q+r)+p(q-r)}{(q+r)(p+q)}$$$$\Rightarrow v=\mathrm{\angle }EFD={\cos }^{-1}\left[\frac{q(q+r)+p(q-r)}{(q+r)(p+q)}\right]$$$$\Rightarrow \cos \mathrm{\angle }FDE=\frac{r(r+p)+q(r-p)}{(r+p)(q+r)}$$$$\Rightarrow w=\mathrm{\angle }FDE={\cos }^{-1}\left[\frac{r(r+p)+q(r-p)}{(r+p)(q+r)}\right]$$$$$$$$\mathrm{\angle }B=u-\alpha +\gamma ={\cos }^{-1}\left[\frac{p(p+q)+r(p-q)}{(p+q)(r+p)}\right]-{\sin }^{-1}\left(\frac{p-q}{p+q}\right)+{\sin }^{-1}\left(\frac{r-p}{r+p}\right)$$$$\mathrm{\angle }C=v-\beta +\alpha ={\cos }^{-1}\left[\frac{q(q+r)+p(q-r)}{(q+r)(p+q)}\right]-{\sin }^{-1}\left(\frac{q-r}{q+r}\right)+{\sin }^{-1}\left(\frac{p-q}{p+q}\right)$$$$\mathrm{\angle }A=w-\gamma +\beta ={\cos }^{-1}\left[\frac{r(r+p)+q(r-p)}{(r+p)(q+r)}\right]-{\sin }^{-1}\left(\frac{r-p}{r+p}\right)+{\sin }^{-1}\left(\frac{q-r}{q+r}\right)$$$$$$$$a=\frac{p}{\tan \frac{B}{2}}+\sqrt{{(p+q)}^2-{(p-q)}^2}+\frac{q}{\tan \frac{C}{2}}$$$$\Rightarrow a=\frac{p(1+\cos B)}{\sin B}+2\sqrt{pq}+\frac{q(1+\cos C)}{\sin C}$$$$\Rightarrow b=\frac{q(1+\cos C)}{\sin C}+2\sqrt{qr}+\frac{r(1+\cos A)}{\sin A}$$$$\Rightarrow c=\frac{r(1+\cos A)}{\sin A}+2\sqrt{rp}+\frac{p(1+\cos B)}{\sin B}$$

Commented by ajfour last updated on 06/Feb/18

$$AmazingSir,thankyou!$$

Commented by mrW2 last updated on 06/Feb/18

Terms of Service

Privacy Policy

Contact: info@tinkutara.com