Exercise (11.6) - Solution


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

Show/Hide Solution
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.$

Show/Hide Solution
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.$

Show/Hide Solution
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.$

Show/Hide Solution
ႏွ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.

Show/Hide Solution
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။
triangle

Triangle solver
Variable Input Solution Solution 2
Area (square units):
Status:


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.

Show/Hide Solution
$ \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 $\displaystyle x = 5.$

Show/Hide Solution
$ \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}$


အေျဖမ်ားကို ဆက္လက္ တင္ေပးပါမည္။
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!
أحدث أقدم