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In-a-triangle-the-length-of-the-two-larger-sides-are-24-and-22-respectively-If-the-angles-are-in-AP-then-the-third-side-is-

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Question Number 41079 by nadeem last updated on 01/Aug/18
In a triangle the length of the two larger sides are 24 and 22, respectively. If the angles are in AP, then the third side is
$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{larger}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{24}\:\mathrm{and}\:\mathrm{22},\:\mathrm{respectively}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{third} \\ $$$$\mathrm{side}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18
let angles are A−∝,A,A+∝ A−∝+A+A+∝=180^o 3A=180^o A=60^o unknoen side be x cosA=((x^2 +24^2 −22^2 )/(2x.24)) (1/2)=((x^2 +46×2)/(2.x.24)) 24x=x^2 +92 x^2 −24x+92=0 x=((24−(√(576−368)) )/2)=((24−14.42)/2)=((9.58)/2)=4.79 we can not take ((24+(√(576−368)) )/2) since 24 is greatest side as per question cos(A−∝)=((24^2 +22^2 −4.79^2 )/(2.24.22)) A−∝=cos^(−1) (((24^2 +22^2 −4.79^2 )/(2.24.22))) A+.∝=cos^(−1) (((22^2 +4.79^2 −24^2 )/(2.22.4.79)))
$${let}\:{angles}\:{are}\:{A}−\propto,{A},{A}+\propto \\ $$$${A}−\propto+{A}+{A}+\propto=\mathrm{180}^{{o}} \\ $$$$\mathrm{3}{A}=\mathrm{180}^{{o}} \:\:\:{A}=\mathrm{60}^{{o}} \\ $$$${unknoen}\:{side}\:{be}\:{x} \\ $$$${cosA}=\frac{{x}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} −\mathrm{22}^{\mathrm{2}} }{\mathrm{2}{x}.\mathrm{24}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{{x}^{\mathrm{2}} +\mathrm{46}×\mathrm{2}}{\mathrm{2}.{x}.\mathrm{24}} \\ $$$$\mathrm{24}{x}={x}^{\mathrm{2}} +\mathrm{92} \\ $$$${x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{92}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{24}−\sqrt{\mathrm{576}−\mathrm{368}}\:}{\mathrm{2}}=\frac{\mathrm{24}−\mathrm{14}.\mathrm{42}}{\mathrm{2}}=\frac{\mathrm{9}.\mathrm{58}}{\mathrm{2}}=\mathrm{4}.\mathrm{79} \\ $$$${we}\:{can}\:{not}\:{take}\:\frac{\mathrm{24}+\sqrt{\mathrm{576}−\mathrm{368}}\:}{\mathrm{2}}\:{since} \\ $$$$\mathrm{24}\:{is}\:{greatest}\:{side}\:{as}\:{per}\:{question} \\ $$$${cos}\left({A}−\propto\right)=\frac{\mathrm{24}^{\mathrm{2}} +\mathrm{22}^{\mathrm{2}} −\mathrm{4}.\mathrm{79}^{\mathrm{2}} }{\mathrm{2}.\mathrm{24}.\mathrm{22}} \\ $$$${A}−\propto={cos}^{−\mathrm{1}} \left(\frac{\mathrm{24}^{\mathrm{2}} +\mathrm{22}^{\mathrm{2}} −\mathrm{4}.\mathrm{79}^{\mathrm{2}} }{\mathrm{2}.\mathrm{24}.\mathrm{22}}\right) \\ $$$${A}+.\propto={cos}^{−\mathrm{1}} \left(\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{4}.\mathrm{79}^{\mathrm{2}} −\mathrm{24}^{\mathrm{2}} }{\mathrm{2}.\mathrm{22}.\mathrm{4}.\mathrm{79}}\right) \\ $$$$ \\ $$$$ \\ $$

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