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Question Number 16392 by ajfour last updated on 21/Jun/17

Answered by Tinkutara last updated on 21/Jun/17

Commented by Tinkutara last updated on 21/Jun/17

∠AMI = 90°, ∠MAI = (A/2), s = ((a + b + c)/2), Δ = (√(s(s − a)(s − b)(s − c))) In ΔAMI, sin ((A/2)) = ((IM)/(AI)) = (Δ/(sx)) ⇒ AI = (Δ/s) cosec ((A/2))

$$\angle{AMI}\:=\:\mathrm{90}°,\:\angle{MAI}\:=\:\frac{{A}}{\mathrm{2}},\:{s}\:=\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}}, \\ $$$$\Delta\:=\:\sqrt{{s}\left({s}\:−\:{a}\right)\left({s}\:−\:{b}\right)\left({s}\:−\:{c}\right)} \\ $$$$\mathrm{In}\:\Delta{AMI},\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right)\:=\:\frac{{IM}}{{AI}}\:=\:\frac{\Delta}{{sx}} \\ $$$$\Rightarrow\:{AI}\:=\:\frac{\Delta}{{s}}\:\mathrm{cosec}\:\left(\frac{{A}}{\mathrm{2}}\right) \\ $$

Commented by Tinkutara last updated on 21/Jun/17

Commented by Tinkutara last updated on 21/Jun/17

[ABC] = [ABP] + [APC] (1/2)bc sin A = (1/2) c.AP sin ((A/2)) + (1/2)b.AP sin ((A/2)) bc sin A = AP (b + c) sin ((A/2)) AP = ((bc sin A)/((b + c) sin ((A/2)))) = ((2bc)/(b + c)) cos ((A/2)) ∴ IP = AP − AI can easily be calculated.

$$\left[{ABC}\right]\:=\:\left[{ABP}\right]\:+\:\left[{APC}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{bc}\:\mathrm{sin}\:{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{c}.{AP}\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$+\:\frac{\mathrm{1}}{\mathrm{2}}{b}.{AP}\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$${bc}\:\mathrm{sin}\:{A}\:=\:{AP}\:\left({b}\:+\:{c}\right)\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$${AP}\:=\:\frac{{bc}\:\mathrm{sin}\:{A}}{\left({b}\:+\:{c}\right)\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}{bc}}{{b}\:+\:{c}}\:\mathrm{cos}\:\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\therefore\:\boldsymbol{{IP}}\:=\:\boldsymbol{{AP}}\:−\:\boldsymbol{{AI}}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{be} \\ $$$$\mathrm{calculated}. \\ $$

Commented by ajfour last updated on 21/Jun/17

very good, thanks .

$${very}\:{good},\:{thanks}\:. \\ $$

Answered by mrW1 last updated on 21/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

hello mrW1. please say me:you draw this perfect diagram by what?thanks.

$${hello}\:{mrW}\mathrm{1}.\:{please}\:{say}\:{me}:{you}\:{draw}\: \\ $$$${this}\:{perfect}\:{diagram}\:{by}\:{what}?{thanks}. \\ $$

Commented by mrW1 last updated on 22/Jun/17

I use Geometry Expression, but on PC.

$$\mathrm{I}\:\mathrm{use}\:\mathrm{Geometry}\:\mathrm{Expression},\:\mathrm{but}\:\mathrm{on}\:\mathrm{PC}. \\ $$

Commented by mrW1 last updated on 21/Jun/17

let s=((a+b+c)/2) let Δ=area of ΔABC=(√(s(s−a)(s−b)(s−c))) u+v=c ...i v+w=a ...ii w+u=b ...iii sum of all: 2(u+v+w)=(a+b+c)=2s ⇒u+v+w=s ...iv (iv)−(i)⇒w=s−c (iv)−(ii)⇒u=s−a (iv)−(iii)⇒v=s−b x^2 =AI^2 =u^2 +r^2 =(s−a)^2 +[(Δ/s)]^2 =(s−a)^2 +((s(s−a)(s−b)(s−c))/s^2 ) =(s−a)[(s−a)+(((s−b)(s−c))/s)] =(s−a)[((2s^2 −s(a+b+c)+bc)/s)] =(s−a)[((2s^2 −2s^2 +bc)/s)] =(((s−a)bc)/s) ⇒x=(√(((s−a)bc)/s)) height of ΔABC over side BC= h_A (1/2)×a×h_A =Δ ⇒h_A =((2Δ)/a) r=(Δ/s) ⇒(h_A /r)=((2s)/a) ((AD)/(ID))=(h_A /r) ⇒((x+y)/y)=(h_A /r)=((2s)/a) ax+ay=2sy (2s−a)y=ax (b+c)y=ax ⇒y=(a/(b+c))x ⇒y=(a/(b+c))(√(((s−a)bc)/s))

$$\mathrm{let}\:\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$$$\mathrm{let}\:\Delta=\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{ABC}=\sqrt{\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$ \\ $$$$\mathrm{u}+\mathrm{v}=\mathrm{c}\:\:...\mathrm{i} \\ $$$$\mathrm{v}+\mathrm{w}=\mathrm{a}\:\:\:...\mathrm{ii} \\ $$$$\mathrm{w}+\mathrm{u}=\mathrm{b}\:\:\:...\mathrm{iii} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{all}: \\ $$$$\mathrm{2}\left(\mathrm{u}+\mathrm{v}+\mathrm{w}\right)=\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=\mathrm{2s} \\ $$$$\Rightarrow\mathrm{u}+\mathrm{v}+\mathrm{w}=\mathrm{s}\:\:\:...\mathrm{iv} \\ $$$$\left(\mathrm{iv}\right)−\left(\mathrm{i}\right)\Rightarrow\mathrm{w}=\mathrm{s}−\mathrm{c} \\ $$$$\left(\mathrm{iv}\right)−\left(\mathrm{ii}\right)\Rightarrow\mathrm{u}=\mathrm{s}−\mathrm{a} \\ $$$$\left(\mathrm{iv}\right)−\left(\mathrm{iii}\right)\Rightarrow\mathrm{v}=\mathrm{s}−\mathrm{b} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{AI}^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} =\left(\mathrm{s}−\mathrm{a}\right)^{\mathrm{2}} +\left[\frac{\Delta}{\mathrm{s}}\right]^{\mathrm{2}} \\ $$$$=\left(\mathrm{s}−\mathrm{a}\right)^{\mathrm{2}} +\frac{\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)}{\mathrm{s}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{s}−\mathrm{a}\right)\left[\left(\mathrm{s}−\mathrm{a}\right)+\frac{\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)}{\mathrm{s}}\right] \\ $$$$=\left(\mathrm{s}−\mathrm{a}\right)\left[\frac{\mathrm{2s}^{\mathrm{2}} −\mathrm{s}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)+\mathrm{bc}}{\mathrm{s}}\right] \\ $$$$=\left(\mathrm{s}−\mathrm{a}\right)\left[\frac{\mathrm{2s}^{\mathrm{2}} −\mathrm{2s}^{\mathrm{2}} +\mathrm{bc}}{\mathrm{s}}\right] \\ $$$$=\frac{\left(\mathrm{s}−\mathrm{a}\right)\mathrm{bc}}{\mathrm{s}} \\ $$$$\Rightarrow\mathrm{x}=\sqrt{\frac{\left(\mathrm{s}−\mathrm{a}\right)\mathrm{bc}}{\mathrm{s}}} \\ $$$$ \\ $$$$\mathrm{height}\:\mathrm{of}\:\Delta\mathrm{ABC}\:\mathrm{over}\:\mathrm{side}\:\mathrm{BC}=\:\mathrm{h}_{\mathrm{A}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{a}×\mathrm{h}_{\mathrm{A}} =\Delta \\ $$$$\Rightarrow\mathrm{h}_{\mathrm{A}} =\frac{\mathrm{2}\Delta}{\mathrm{a}} \\ $$$$\mathrm{r}=\frac{\Delta}{\mathrm{s}} \\ $$$$\Rightarrow\frac{\mathrm{h}_{\mathrm{A}} }{\mathrm{r}}=\frac{\mathrm{2s}}{\mathrm{a}} \\ $$$$ \\ $$$$\frac{\mathrm{AD}}{\mathrm{ID}}=\frac{\mathrm{h}_{\mathrm{A}} }{\mathrm{r}} \\ $$$$\Rightarrow\frac{\mathrm{x}+\mathrm{y}}{\mathrm{y}}=\frac{\mathrm{h}_{\mathrm{A}} }{\mathrm{r}}=\frac{\mathrm{2s}}{\mathrm{a}} \\ $$$$\mathrm{ax}+\mathrm{ay}=\mathrm{2sy} \\ $$$$\left(\mathrm{2s}−\mathrm{a}\right)\mathrm{y}=\mathrm{ax} \\ $$$$\left(\mathrm{b}+\mathrm{c}\right)\mathrm{y}=\mathrm{ax} \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}\mathrm{x} \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}\sqrt{\frac{\left(\mathrm{s}−\mathrm{a}\right)\mathrm{bc}}{\mathrm{s}}} \\ $$

Commented by ajfour last updated on 22/Jun/17

thank you sir, very natural, loved reading the solution..

$$\:{thank}\:{you}\:{sir},\:{very}\:{natural},\:{loved} \\ $$$$\:{reading}\:{the}\:{solution}.. \\ $$

Commented by mrW1 last updated on 22/Jun/17

thank you too, sir! At first I always try to solve without using trigonomical formula which I can′t remember well.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{too},\:\mathrm{sir}! \\ $$$$\mathrm{At}\:\mathrm{first}\:\mathrm{I}\:\mathrm{always}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{without} \\ $$$$\mathrm{using}\:\mathrm{trigonomical}\:\mathrm{formula}\:\mathrm{which}\:\mathrm{I} \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{remember}\:\mathrm{well}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

((BD)/(DC))=(c/b)⇒((BD+DC)/(DC))=((b+c)/b)⇒DC=((ab)/(b+c)) ⇒ { ((BD=((ac)/(b+c)))),((DC=((ab)/(b+c)))) :} a(AD^2 +BD.DC)=b^2 .BD+c^2 .DC⇒ a(AD^2 +((a^2 bc)/((b+c)^2 )))=((acb^2 )/(b+c))+((abc^2 )/(b+c))(Estewart teorem) ⇒AD^2 +((a^2 bc)/((b+c)^2 ))= bc⇒AD^2 =bc−((a^2 bc)/((b+c)^2 )) AD^2 =bc((((b+c)^2 −a^2 )/((b+c)^2 )))=bc((2p.2(p−a))/((b+c)^2 )) ⇒AD=(2/(b+c))(√(bcp(p−a)))=((2bc)/(b+c)).cos(A/2) ((AI)/(sin(B/2)))=(c/(sin(180−((A+B)/2))))=(c/(cos(C/2)))⇒ AI=c.sin(B/2).sec(C/2)=c(√(((p−a)(p−c))/(ac))).(√((ba)/(p(p−c)))) ⇒AI=x=(√(bc.((p−a)/p))) (y/(sin(C/2)))=((DC)/(sin((A+C)/2)))=((DC)/(cos(B/2))) y=DC.sin(C/2).sec(B/2)=((ab)/(b+c))(√(((p−a)(p−b))/(ab))).(√((ac)/(p(p−b))))= =(a/(b+c))(√(bc.(((p−a))/p))) .■ note: ID=(a/(b+c))(√(bc.((p−a)/p))),AI=(√(bc.((p−a)/p))) IE=(b/(a+c))(√(ac.((p−b)/p))),BI=(√(ac.((p−b)/p))) IF=(c/(a+b))(√(ab.((p−c)/p))),CI=(√(ab.((p−c)/p)))

$$\frac{{BD}}{{DC}}=\frac{{c}}{{b}}\Rightarrow\frac{{BD}+{DC}}{{DC}}=\frac{{b}+{c}}{{b}}\Rightarrow{DC}=\frac{{ab}}{{b}+{c}} \\ $$$$\Rightarrow\begin{cases}{{BD}=\frac{{ac}}{{b}+{c}}}\\{{DC}=\frac{{ab}}{{b}+{c}}}\end{cases} \\ $$$${a}\left({AD}^{\mathrm{2}} +{BD}.{DC}\right)={b}^{\mathrm{2}} .{BD}+{c}^{\mathrm{2}} .{DC}\Rightarrow \\ $$$${a}\left({AD}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {bc}}{\left({b}+{c}\right)^{\mathrm{2}} }\right)=\frac{{acb}^{\mathrm{2}} }{{b}+{c}}+\frac{{abc}^{\mathrm{2}} }{{b}+{c}}\left({Estewart}\:{teorem}\right) \\ $$$$\Rightarrow{AD}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {bc}}{\left({b}+{c}\right)^{\mathrm{2}} }=\:{bc}\Rightarrow{AD}^{\mathrm{2}} ={bc}−\frac{{a}^{\mathrm{2}} {bc}}{\left({b}+{c}\right)^{\mathrm{2}} } \\ $$$${AD}^{\mathrm{2}} ={bc}\left(\frac{\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }{\left({b}+{c}\right)^{\mathrm{2}} }\right)={bc}\frac{\mathrm{2}{p}.\mathrm{2}\left({p}−{a}\right)}{\left({b}+{c}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{AD}=\frac{\mathrm{2}}{{b}+{c}}\sqrt{{bcp}\left({p}−{a}\right)}=\frac{\mathrm{2}{bc}}{{b}+{c}}.{cos}\frac{{A}}{\mathrm{2}} \\ $$$$\frac{{AI}}{{sin}\frac{{B}}{\mathrm{2}}}=\frac{{c}}{{sin}\left(\mathrm{180}−\frac{{A}+{B}}{\mathrm{2}}\right)}=\frac{{c}}{{cos}\frac{{C}}{\mathrm{2}}}\Rightarrow \\ $$$${AI}={c}.{sin}\frac{{B}}{\mathrm{2}}.{sec}\frac{{C}}{\mathrm{2}}={c}\sqrt{\frac{\left({p}−{a}\right)\left({p}−{c}\right)}{{ac}}}.\sqrt{\frac{{ba}}{{p}\left({p}−{c}\right)}} \\ $$$$\Rightarrow{AI}={x}=\sqrt{{bc}.\frac{{p}−{a}}{{p}}} \\ $$$$\frac{{y}}{{sin}\frac{{C}}{\mathrm{2}}}=\frac{{DC}}{{sin}\frac{{A}+{C}}{\mathrm{2}}}=\frac{{DC}}{{cos}\frac{{B}}{\mathrm{2}}} \\ $$$${y}={DC}.{sin}\frac{{C}}{\mathrm{2}}.{sec}\frac{{B}}{\mathrm{2}}=\frac{{ab}}{{b}+{c}}\sqrt{\frac{\left({p}−{a}\right)\left({p}−{b}\right)}{{ab}}}.\sqrt{\frac{{ac}}{{p}\left({p}−{b}\right)}}= \\ $$$$=\frac{{a}}{{b}+{c}}\sqrt{{bc}.\frac{\left({p}−{a}\right)}{{p}}}\:\:\:\:.\blacksquare \\ $$$${note}: \\ $$$${ID}=\frac{{a}}{{b}+{c}}\sqrt{{bc}.\frac{{p}−{a}}{{p}}},{AI}=\sqrt{{bc}.\frac{{p}−{a}}{{p}}} \\ $$$${IE}=\frac{{b}}{{a}+{c}}\sqrt{{ac}.\frac{{p}−{b}}{{p}}},{BI}=\sqrt{{ac}.\frac{{p}−{b}}{{p}}} \\ $$$${IF}=\frac{{c}}{{a}+{b}}\sqrt{{ab}.\frac{{p}−{c}}{{p}}},{CI}=\sqrt{{ab}.\frac{{p}−{c}}{{p}}} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

great thank to you dear mr Ajfour.

$${great}\:{thank}\:{to}\:{you}\:{dear}\:{mr}\:{Ajfour}. \\ $$

Commented by ajfour last updated on 22/Jun/17

I am very grateful to you sir, utterly complete an analysis this is.

$${I}\:{am}\:{very}\:{grateful}\:{to}\:{you}\:{sir}, \\ $$$$\:{utterly}\:{complete}\:{an}\:{analysis}\:{this}\:{is}. \\ $$

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