Question Number 206885 by efronzo1 last updated on 29/Apr/24
Answered by BaliramKumar last updated on 29/Apr/24
$${x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)\:\approx\:\mathrm{32}° \\ $$
Answered by Rasheed.Sindhi last updated on 29/Apr/24
![Four triangles+smaller square=large square ((a(a+3))/2)×4+9=89 [a is smaller side of a triangle] 2a^2 +6a−80=0 a^2 +3a−40=0 (a+8)(a−5)=0 a≠−8⇒a=5 tan x=(5/8) x=tan^(−1) ((5/8))≈32°](https://www.tinkutara.com/question/Q206887.png)
$$\mathrm{Four}\:\mathrm{triangles}+\mathrm{smaller}\:\mathrm{square}=\mathrm{large}\:\mathrm{square} \\ $$$$\frac{{a}\left({a}+\mathrm{3}\right)}{\mathrm{2}}×\mathrm{4}+\mathrm{9}=\mathrm{89}\:\left[{a}\:{is}\:{smaller}\:{side}\:{of}\:{a}\:{triangle}\right] \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}{a}−\mathrm{80}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{a}−\mathrm{40}=\mathrm{0} \\ $$$$\left({a}+\mathrm{8}\right)\left({a}−\mathrm{5}\right)=\mathrm{0} \\ $$$${a}\neq−\mathrm{8}\Rightarrow{a}=\mathrm{5} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)\approx\mathrm{32}° \\ $$
Answered by Rasheed.Sindhi last updated on 29/Apr/24
![Let smaller side of triangle is a a^2 +(a+3)^2 =89 [Pythagoras theorm] 2a^2 +6a+9−89=0 a^2 +3a+40=0 (a+8)(a−5)=0 a≠−8⇒a=5 tan x=(a/(a+3))=(5/8) x=tan^(−1) ((5/8))=32°](https://www.tinkutara.com/question/Q206889.png)
$${Let}\:{smaller}\:{side}\:{of}\:{triangle}\:{is}\:{a} \\ $$$${a}^{\mathrm{2}} +\left({a}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{89}\:\left[\mathrm{Pythagoras}\:\mathrm{theorm}\right] \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{9}−\mathrm{89}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{40}=\mathrm{0} \\ $$$$\left({a}+\mathrm{8}\right)\left({a}−\mathrm{5}\right)=\mathrm{0} \\ $$$${a}\neq−\mathrm{8}\Rightarrow{a}=\mathrm{5} \\ $$$$\mathrm{tan}\:{x}=\frac{{a}}{{a}+\mathrm{3}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)=\mathrm{32}° \\ $$
Answered by mr W last updated on 30/Apr/24
$${s}^{\mathrm{2}} =\mathrm{89} \\ $$$$\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}×{s}^{\mathrm{2}} \mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{89}−\mathrm{9}=\mathrm{80} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\frac{\mathrm{80}}{\mathrm{89}}\: \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{80}}{\mathrm{89}}\approx\mathrm{32}° \\ $$