Exercise (11.6) - Solution


1.        Solve $ \displaystyle \vartriangle ABC$ with $ \displaystyle b = 18.1, c = 12.3,$ and $ \displaystyle a=115{}^\circ .$

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11.6_1_inline


$ \displaystyle \begin{array}{l}\ \ \ \ \ \ b=18.1,\ c=12.3\ \text{and}\ \alpha =115{}^\circ \\\\ \ \ \ \ \ \ \text{By the law of cosines,}\\\\ \ \ \ \ \ \ {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos \alpha \\\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{(18.1)}^{2}}+{{(12.3)}^{2}}-2(18.1)(12.3)\cos 115{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ =327.6\text{ }+151.3-\left( {36.2} \right)\text{ }\left( {12.3} \right)\cos (180{}^\circ -65{}^\circ )\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =478.9+\left( {36.2} \right)\text{ }\left( {12.3} \right)\cos 65{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 36.2} & {1.5587} \\ { 12.3 } & {1.0899}\\ {\cos65{}^\circ} & {\overline{1}.6259} \\ \hline {188.1} & {2.2745} \\ \hline \end{array}\end{array}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =478.9+188.1\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =667\\\\ \ \ \ \ \ \ \ \ a\ \ =\sqrt{{667}}=25.83\\\\ \ \ \ \ \ \ \text{By the law of sines,}\ \\\\ \ \ \ \ \ \ \cos \gamma =\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{667+327.6-151.3}}{{2(25.83)(18.1)}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{843.3}}{{(25.83)(36.2)}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 843.3} & {2.9260} \\ { 25.83 } & {1.4121}\\ {36.2} & {1.5587} \\ \hline {} & {2.9708} \\ \hline {\cos 25 {}^\circ{35}'} & {\overline{1}.9552} \\ \hline \end{array}\end{array}\\\\ \ \ \ \ \ \ \ \ \ \ \ \gamma \ \ =25{}^\circ 3{5}'\\\\\therefore ~~~\ \ \beta =\text{ }180{}^\circ -\left( {\alpha +\gamma } \right)\\\\ \ \ \ \ \ \ \ \ \ \ =180{}^\circ -(115{}^\circ +\text{ }25{}^\circ 3{5}')\\\\ \ \ \ \ \ \ \ \ \ \ =180{}^\circ -140{}^\circ 3{5}'\\\\ \ \ \ \ \ \ \ \ \ \ =39{}^\circ 2{5}'\end{array}$


2.        In $ \displaystyle \vartriangle ABC, a = 5, \beta = 75{}^\circ$ and $ \displaystyle \gamma = 41{}^\circ.$ Find $ \displaystyle \alpha, b$ and $ \displaystyle c.$

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11.6_2_inline

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ a=5,\ \beta =75{}^\circ \ \text{and}\ \gamma =41{}^\circ \\\\ \therefore \ \ \ \ \alpha =\text{ }180{}^\circ -(\beta +\gamma )\\\\ \ \ \ \ \ \ \ \ \ \ =180{}^\circ -(75{}^\circ +\text{ }41{}^\circ )\\\\ \ \ \ \ \ \ \ \ \ \ =180{}^\circ -116{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ =64{}^\circ \\\\ \ \ \ \ \ \ \text{By the law of sines,}\\\\ \ \ \ \ \ \ \ \displaystyle \frac{b}{{\sin \beta }}=\displaystyle \frac{a}{{\sin \alpha }}\\\\ \therefore \ \ \ \ b=\displaystyle \frac{{a\sin \beta }}{{\sin \alpha }}\\\\ \therefore \ \ \ \ \ b=a\sin \beta \operatorname{cosec}\alpha \\\\ \ \ \ \ \ \ \ \ \ \ =5\text{ }\sin 75{}^\circ \operatorname{cosec}64{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 5} & {0.6990} \\ { \sin 75 {}^\circ } & {\overline{1}.9849}\\ { \displaystyle \operatorname{cosec}64{}^\circ } & {0.0463} \\ \hline {5.373} & {0.7302} \\ \hline \end{array}\end{array}\\\\ \ \ \ \ \ \ \ \ \ \ =5.373\\\\\ \ \ \ \ \ \ \text{Similarly,}\ \\\\ \ \ \ \ \ \ \ \displaystyle \frac{c}{{\sin \gamma }}=\displaystyle \frac{a}{{\sin \alpha }}\\\\ \therefore \ \ \ \ \ \ c=\displaystyle \frac{{a\sin \gamma }}{{\sin \alpha }}\\\\ \therefore \ \ \ \ \ \ c=a\sin \gamma \operatorname{cosec}\alpha \\\\ \ \ \ \ \ \ \ \ \ \ =5\text{ }\sin 41{}^\circ \operatorname{cosec}64{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 5} & {0.6990} \\ { \sin 41 {}^\circ } & {\overline{1}.8169}\\ { \displaystyle \operatorname{cosec}64{}^\circ } & {0.0463} \\ \hline {3.65} & {0.5622} \\ \hline \end{array}\end{array}\\\\ \ \ \ \ \ \ \ \ \ \ =3.65\end{array}$


3.        Solve $ \displaystyle \vartriangle ABC$ with $ \displaystyle a = 3, b = 4, c = 6.$

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11.6_3

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ a=3,\ b=4,\ c=6\\\\ \ \ \ \ \ \ \text{By the law of cosines},\\\\ \ \ \ \ \ \ \cos \alpha =\displaystyle \frac{{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}{{2bc}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{4}^{2}}+{{6}^{2}}-{{3}^{2}}}}{{2(4)(6)}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{43}}{{48}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 43} & {1.6335} \\ { 48 } & {1.6812}\\ \hline {\cos 26 {}^\circ2{3}'} & {\overline{1}.9523} \\ \hline \end{array}\end{array}\\\\ \therefore \ \ \ \ \cos \alpha =\cos 26{}^\circ 2{3}'\\\\\therefore \ \ \ \ \alpha =26{}^\circ 2{3}'\\\\ \ \ \ \ \ \ \text{Similarly,}\\\\ \ \ \ \ \ \ \cos \beta =\displaystyle \frac{{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}}{{2ac}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{3}^{2}}+{{6}^{2}}-{{4}^{2}}}}{{2(3)(6)}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{29}}{{36}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 29} & {1.4624} \\ { 36 } & {1.5563}\\ \hline {\cos 36 {}^\circ2{0}'} & {\overline{1}.9061} \\ \hline \end{array}\end{array}\\\\ \therefore \ \ \ \ \cos \beta =\cos 36{}^\circ 2{0}'\\\\ \therefore \ \ \ \ \beta =36{}^\circ 2{0}'\\\\ \therefore \ \ \ \ \gamma =180{}^\circ -(\alpha +\beta )\\\\ \therefore \ \ \ \ \gamma =180{}^\circ -(26{}^\circ 2{3}'\text{ }+\text{ }36{}^\circ 2{0}')\\\\ \therefore \ \ \ \ \gamma =117{}^\circ 1{7}'\end{array}$


4.        Solve $ \displaystyle \vartriangle ABC$ with $ \displaystyle a = 3, b = 4, \beta = 50{}^\circ.$

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ႏွစ္နားႏွင့္ မ်က္ႏွာခ်င္ဆိုင္ေထာင့္ ေပးထားေသာေၾကာင့္ ျဖစ္ႏိုင္ေသာ ႀတိဂံအေရအတြက္ကို စဥ္းစားေပးရမည္။

၎ကို Ambiguous Case ဟု ေခၚသည္။

ေပးထားေသာ အေျခေနမ်ားကို လိုက္၍

(1) ႀတိဂံမျဖစ္ႏိုင္ျခင္း

(2) ႀတိဂံတစ္ခုသာ ျဖစ္ႏိုင္ျခင္း

(3) ႀတိဂံ ႏွစ္ခု ျဖစ္ႏိုင္ျခင္း ဟူ၍ အေျခအေနသံုးမ်ိဳးရွိသည္။

ေအာက္ပါဇယားကို ေလ့လာၾကည့္ပါ။

Ambiguous Case

ယခုပုစာၦတြင္ $ \displaystyle \alpha$ ႏွင့္မ်က္ႏွာခ်င္းဆိုင္ အနား $ \displaystyle a$ သည္ ေပးထားေသာ အနား $ \displaystyle b$ ထက္ႀကီးေသာေၾကာင့္ ျဖစ္ႏိုင္ေသာ ႀတိဂံ တစ္ခုသာ ရွိသည္။

11.6_4

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ a=3,\ \ b=4,\ \ \beta =50{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \text{By the law of sines,}\ \\\\ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{\sin \alpha }}{a}=\displaystyle \frac{{\sin \beta }}{b}\\\\ \therefore \ \ \ \ \ \ \ \sin \alpha =\displaystyle \frac{{a\sin \beta }}{b}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{3\sin 50{}^\circ }}{4}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{4}(0.7660)\\\\ \therefore \ \ \ \ \ \ \ \alpha =35{}^\circ {4}'\\\\ \therefore \ \ \ \ \ \ \ \gamma =180{}^\circ -(\alpha +\beta )\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =180{}^\circ -(35{}^\circ {4}'+50{}^\circ )\\\\ \ \ \ \ \ \ \ \ \ \text{Again,}\\\\ \ \ \ \ \ \ \ \ \ \displaystyle \frac{c}{{\sin \gamma }}=\displaystyle \frac{b}{{\sin \beta }}\\\\ \therefore \ \ \ \ \ \ \ \ c=\displaystyle \frac{{b\sin \gamma }}{{\sin \beta }}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =b\sin \gamma \operatorname{cosec}\beta \\\\ \ \ \ \ \ \ \ \ \ \ \ \ =4\sin 94{}^\circ {4}'\operatorname{cosec}50{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ =4\sin (180{}^\circ -94{}^\circ {4}')\operatorname{cosec}50{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ =4\sin 85{}^\circ 5{6}'\operatorname{cosec}50{}^\circ \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 4} & {0.6021} \\ { \sin 85 {}^\circ5{6}' } & {\overline{1}.9989}\\ { \operatorname{cosec}50{}^\circ } & {0.1157} \\ \hline {5.201} & {0.7167} \\ \hline \end{array}\end{array}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =5.201\end{array}$


5.        In $ \displaystyle \vartriangle ABC, a = 15, b = 20, \alpha = 30{}^\circ.$ Solve the triangle.

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ယခုပုစာၦသည္လည္း No(4) ကဲ့သို႔ ႏွစ္နားႏွင့္ မ်က္ႏွာခ်င္းဆိုင္ေထာင့္ ေပးထားရာ ျဖစ္ႏိုင္ေသာ ႀတိဂံအေရအတြက္ကို ဆံုးျဖတ္ေပးရမည္။

ေအာက္ပါ ဇယားျဖင့္ ဆံုးျဖတ္ပါမည္။

Ambiguous Case

$ \displaystyle \alpha$ ႏွင့္မ်က္ႏွာခ်င္းဆိုင္ အနား $ \displaystyle a$ သည္ ေပးထားေသာ အနား $ \displaystyle b$ ေအာက္ငယ္သည္။

ထို႔ေၾကာင့္ $ \displaystyle b\sin \alpha$ ကို စစ္ေဆးပါမည္။

$ \displaystyle b\sin \alpha=20 \sin 30{}^\circ=20\frac{1}{2}=10$

ထို႔ေၾကာင့္ $ \displaystyle b\sin \alpha$< $ \displaystyle a$<$ \displaystyle b$ ျဖစ္ေနေသာေၾကာင့္ ျဖစ္ႏိုင္ေသာ ႀတိဂံ ႏွစ္ခုရွိသည္။

11.6_5

အမွတ္ $ \displaystyle C$ မွ အခ်င္း၀က္ $ \displaystyle 15$ ယူနစ္ရွိေသာ အ၀န္းပိုင္းကို ဆြဲလိုက္သည့္ အခါ မ်ဥ္းျပတ္ $ \displaystyle AM$ ကို အမွတ္ႏွစ္ေနရာ ($ \displaystyle B_1$ ႏွင့္ $ \displaystyle B_2$) တြင္ ျဖတ္သြားသည္။

ထို႔ေၾကာင့္ ေပးထားေသာ အခ်က္အလက္မ်ားကို ေျပလည္ေစေသာ ႀတိဂံႏွစ္ခု ရွိသည္။ ၎တို႔မွာ $ \displaystyle \vartriangle A{{B}_{1}}C$ ႏွင့္ $ \displaystyle \vartriangle A{{B}_{2}}C$ တို႔ ျဖစ္ၾကသည္။

ထို႔ေၾကာင့္ ႀတိဂံတစ္ခုစီအတြက္ သက္ဆိုင္ေသာ လိုအပ္ခ်က္မ်ားကို ရွာေပးရပါမည္။

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \text{In }\vartriangle A{{B}_{1}}C\text{, by the law of sines,}\\\ \\\ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{\sin {{\beta }_{1}}}}{b}=\displaystyle \frac{{\sin \alpha }}{a}\\\\\therefore \ \ \ \ \ \ \ \ \sin {{\beta }_{1}}=\displaystyle \frac{{b\sin \alpha }}{a}\\\\\ \ \ \ \ \ \ \ \ \ \sin {{\beta }_{1}}=\displaystyle \frac{{20\sin 30{}^\circ }}{{50}}\\\\\ \ \ \ \ \ \ \ \ \ \sin {{\beta }_{1}}=\displaystyle \frac{{20}}{{50}}\left( {\displaystyle \frac{1}{2}} \right)=0.6667\\\\\therefore \ \ \ \ \ \ \ \ {{\beta }_{1}}=41{}^\circ 4{9}'\\\\\ \ \ \ \ \ \ \ \ \ \text{But }{{\beta }_{1}}\ \text{is obtuse,}\ 41{}^\circ 4{9}'\ \text{is }\\\ \ \ \ \ \ \ \ \ \ \text{the basic acute angle of }{{\beta }_{1}}.\\\\\therefore \ \ \ \ \ \ \ \ {{\beta }_{1}}=180{}^\circ -41{}^\circ 4{9}'=138{}^\circ 1{1}'\\\\\therefore \ \ \ \ \ \ \ \ {{\gamma }_{1}}=180{}^\circ -(30{}^\circ +138{}^\circ 1{1}')=11{}^\circ 4{9}'\\\\\ \ \ \ \ \ \ \ \ \ \text{By the law of sines,}\\\\\ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{{{c}_{1}}}}{{\sin {{\gamma }_{1}}}}=\displaystyle \frac{a}{{\sin \alpha }}\\\\\therefore \ \ \ \ \ \ \ \ {{c}_{1}}=\displaystyle \frac{{a\sin {{\gamma }_{1}}}}{{\sin \alpha }}=a\sin {{\gamma }_{1}}\operatorname{cosec}\alpha \\\\\therefore \ \ \ \ \ \ \ \ {{c}_{1}}=15\sin 11{}^\circ 4{9}'\operatorname{cosec}30{}^\circ \\\\\therefore \ \ \ \ \ \ \ \ {{c}_{1}}=15\left( {0.2048} \right)\left( 2 \right)\\\\\therefore \ \ \ \ \ \ \ \ {{c}_{1}}=6.144\end{array}$

အထက္ပါရွင္းလင္းခ်က္သည္ $ \displaystyle \vartriangle A{{B}_{1}}C$ အတြက္ ျဖစ္သည္။

$ \displaystyle \vartriangle A{{B}_{2}}C$ အတြက္ ဆက္လက္ ေျဖရွင္းေပးရမည္။

ဆိုလိုသည္မွာ $ \displaystyle \vartriangle A{{B}_{2}}C$ ႏွင့္ ဆိုင္ေသာ $ \displaystyle {c}_{2}, {\beta}_{2}, {\gamma}_{2}$ ကို ရွာေပးရမည္။

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \text{Since }\vartriangle C{{B}_{1}}{{B}_{2}}\text{ is isosceles,}\\\ \\\ \ \ \ \ \ \ \ \ \ {{\beta }_{2}}=\angle C{{B}_{1}}{{B}_{2}}\\\\\therefore \ \ \ \ \ \ \ \ {{\beta }_{2}}=180{}^\circ -{{\beta }_{1}}\\\\\ \ \ \ \ \ \ \ \ \ {{\beta }_{2}}=180{}^\circ -138{}^\circ 1{1}'\\\\\therefore \ \ \ \ \ \ \ \ {{\beta }_{2}}=41{}^\circ 4{9}'\\\\\therefore \ \ \ \ \ \ \ \ {{\gamma }_{2}}=180{}^\circ -(\alpha +{{\beta }_{2}})\\\\\ \ \ \ \ \ \ \ \ \ \ {{\gamma }_{2}}=180{}^\circ -(30{}^\circ +41{}^\circ 4{9}')\\\\\therefore \ \ \ \ \ \ \ \ {{\gamma }_{2}}=108{}^\circ 1{1}'\\\\\ \ \ \ \ \ \ \ \ \ \text{In }\vartriangle A{{B}_{2}}\text{C, by the law of sines,}\\\\\ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{{{c}_{2}}}}{{\sin {{\gamma }_{2}}}}=\displaystyle \frac{a}{{\sin \alpha }}\\\\\therefore \ \ \ \ \ \ \ \ {{c}_{2}}=\displaystyle \frac{{a\sin {{\gamma }_{2}}}}{{\sin \alpha }}=a\sin {{\gamma }_{2}}\operatorname{cosec}\alpha \\\\\therefore \ \ \ \ \ \ \ \ {{c}_{2}}=15\sin 108{}^\circ 1{1}'\operatorname{cosec}30{}^\circ \\\\\therefore \ \ \ \ \ \ \ \ {{c}_{2}}=15\sin \left( {180{}^\circ -108{}^\circ 1{1}'} \right)\operatorname{cosec}30{}^\circ \\\ \ \ \ \ \ \ \ \ \ \left[ {\because \sin 108{}^\circ 1{1}'=in\left( {180{}^\circ -108{}^\circ 1{1}'} \right)} \right]\\\\\therefore \ \ \ \ \ \ \ \ {{c}_{2}}=15\sin 71{}^\circ 4{9}'\operatorname{cosec}30{}^\circ \\\\\therefore \ \ \ \ \ \ \ \ {{c}_{2}}=15(0.9500)(2)=28.5\end{array}$


6.        Solve the following triangles.

$\displaystyle \begin{array}{l}\begin{array}{rlllcccccc} \text{(a)} & {\alpha =25{}^\circ } & {\beta =55{}^\circ} & {b=12}\\\\ \text{(b)} & {\gamma =110{}^\circ} & {\beta =28{}^\circ} & {a=8}\\\\ \text{(c)} & {a=9} & {b=11} & {\gamma =60{}^\circ}\\\\ \text{(d)} & {a=5} & {b=8} & {c=7}\\\\ \text{(e)} & {\angle A=154{}^\circ} & {\angle B=15{}^\circ 3{0}'} & {c=20}\\\\ \text{(f)} & {\angle B=64{}^\circ 2{0}'} & {\angle B=50{}^\circ } & {b=5} \end{array}\end{array}$


            (a) မွ (f) အထိပုစာၦမ်ားကို ကိုယ္တိုင္တြက္ ၾကည့္ပါ။ တြက္နည္းမ်ားကို ေနာက္ရက္တြင္ ဆက္လက္တင္ေပးပါမည္။ အေျဖမွန္/မမွန္ကို ေအာက္တြင္ ေပးထားေသာ calculator ျဖင့္ စစ္ေဆးႏိုင္ပါသည္။
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7.        In $ \displaystyle \vartriangle ABC$, $ \displaystyle AB = x, BC = x + 2$, $ \displaystyle AC = x – 2$ where $ \displaystyle x > 4,$ prove that $ \displaystyle \cos A=\frac{{x-8}}{{2(x-2)}}.$ Find the integral values of $ \displaystyle x$ for which $ \displaystyle A$ is obtuse.

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$ \displaystyle \begin{array}{l}\ \ \ \ AB=x,\ BC=x+2,\ AC=x-2,\ x>4\\\\\ \ \ \ \text{By the law of cosines,}\\\\\ \ \ \ \cos A=\displaystyle \frac{{A{{B}^{2}}+A{{C}^{2}}-B{{C}^{2}}}}{{2\cdot AB\cdot AC}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}+{{{(x-2)}}^{2}}-{{{(x+2)}}^{2}}}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}+{{x}^{2}}-4x+4-{{x}^{2}}-4x+4}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}-8x}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x-8}}{{2(x-2)}}\end{array}$


ဒုတိယပိုင္းမွာ $ \displaystyle \angle A$ ရဲ့ ကိန္းျပည့္တန္ဖိုး (integral value) ကို ရွာခိုင္းတယ္။ $ \displaystyle \angle A$ က ေထာင့္က်ယ္ (obtuse) လို႔ ေပးထားတယ္။

obtuse လို႔ ေပးထားလို႔ cartesian plane မွာ $ \displaystyle \angle A$ ရဲ့ လက္တံ (terminal side) က second quadrant မွာ ရွိေနမယ္လို႔ သိရမယ္။

ဒါေၾကာင့္ $ \displaystyle \cos A$ ရဲ့ တန္ဖိုးက အႏႈတ္ကိန္း ျဖစ္မယ္။ ဆိုလိုတာက $ \displaystyle \cos A<0$ ျဖစ္မယ္။

$ \displaystyle \frac{a}{b}<0$ ျဖစ္တဲ့ အခါ a နဲ႔ b မွာ ျဖစ္ႏိုင္တဲ့ လကၡဏာ ေတြဟာ $ \displaystyle ab<0$ မွာ ျဖစ္ႏိုင္တဲ့ ပံုစံနဲ႔ အတူတူပါပဲ။

$ \displaystyle \begin{array}{l}\ \displaystyle \frac{a}{b}<0,\\\\ a<0\ \ \text{(and)}\ b>0\ \ \text{(or)}\ \ a>0\ \ \text{(and)}\ b<0\end{array}$

ယခုေမးခြန္းမွာေတာ့ $ \displaystyle x>4$ လို႔ ေပးထားလို႔ ပိုင္းေျခ $ \displaystyle x-2=4-2=2>0$ ျဖစ္ေနတယ္။

ဒါေၾကာင့္ အထက္မွာ ေျပာခဲ့တဲ့ possibilities ႏွစ္မ်ိဳးထဲက ပထမအေျခအေန ပဲ ျဖစ္ႏိုင္ေတာ့ တယ္။

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Since }\cos A<0,\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{x-8}}{{2(x-2)}}<0\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{x-8}}{{2(x-2)}}\times 2<0\times 2\\\\\therefore \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle \frac{{x-8}}{{x-2}}<0\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{But}\ x>4,\ \ \ \ \left[ {\because \text{given}} \right]\ \text{ }\\\\\therefore \ \ \ \ \ \ \ \ \ \ \ \ x-2=4-2=2>0\\\\\therefore \ \ \ \ \ \ \ \ \ \ \ \ x-8<0\Rightarrow x<8\\\\\therefore \ \ \ \ \ \ \ \ \ \ \ \ 4 < x < 8\\\\\therefore \ \ \ \ \ \ \ \ \ \ \ \ \text{The integral values of }x\text{ are 5, 6 and 7}\text{.}\end{array}$



8.        In $ \displaystyle \vartriangle ABC$, $ \displaystyle AB = x - 2$, $ \displaystyle BC = x$, $ \displaystyle AC = x + 2$ where $ \displaystyle x > 4,$ find the cosine of $\displaystyle \angle ABC.$ Hence find $\displaystyle \angle ABC$ when $\displaystyle x = 5.$

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$ \displaystyle \begin{array}{l}\ \ \ \ AB=x-2,\ BC=x,\ AC=x+2,\ x>4\\\\\ \ \ \ \text{By the law of cosines,}\\\\\ \ \ \ \cos \left( {\angle ABC} \right)=\displaystyle \frac{{A{{B}^{2}}+B{{C}^{2}}-A{{C}^{2}}}}{{2\cdot AB\cdot BC}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{{(x-2)}}^{2}}+{{x}^{2}}-{{{(x+2)}}^{2}}}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}-4x+4+{{x}^{2}}-{{x}^{2}}-4x-4}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}-8x}}{{2\cdot x\cdot (x-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x-8}}{{2(x-2)}}\\\\\ \ \ \ \text{Since }x=5\ \\\\\ \ \ \ \cos \left( {\angle ABC} \right)=\ \displaystyle \frac{{5-8}}{{2(5-2)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ -\displaystyle \frac{1}{2}\\\\\therefore \ \ \ \cos \left( {\angle ABC} \right)=\ -\cos 60{}^\circ \\\\\therefore \ \ \ \cos \left( {\angle ABC} \right)=\ \cos \left( {180{}^\circ -60{}^\circ } \right)\\\\\therefore \ \ \ \cos \left( {\angle ABC} \right)=\ \cos 120{}^\circ \\\\\therefore \ \ \ \angle ABC=\ 120{}^\circ \end{array}$


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