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Question Number 203171 by a.lgnaoui last updated on 11/Jan/24

$$\mathrm{Ratio}\mathrm{of}\mathrm{green}\mathrm{blue}\mathrm{and}\mathrm{Red}\mathrm{triangles}?$$$$\mathrm{Evaluate}\mathrm{these}\mathrm{Area}\mathrm{for}\mathit{\alpha }=25°\mathrm{and}$$$$\mathrm{length}\mathrm{of}\mathrm{aquart}=1$$

Commented by a.lgnaoui last updated on 11/Jan/24

Commented by a.lgnaoui last updated on 11/Jan/24

Answered by Rasheed.Sindhi last updated on 13/Jan/24

Commented by Rasheed.Sindhi last updated on 13/Jan/24

$$\mathrm{Blue}△:$$$$\cos \left(25\right)=\frac{a}{c}=\frac{a}{1}\Rightarrow a=\cos \left(25\right)$$$$\sin \left(25\right)=\frac{b}{c}=\frac{b}{1}\Rightarrow b=\sin \left(25\right)$$$$\mathrm{Blue}▴=\frac{ab}{2}=\frac{\cos \left(25\right)\sin \left(25\right)}{2}$$$$\overline{)\begin{array}{c}\mathrm{Blue}▴=\frac{ab}{2}=m\frac{\cos \left(25\right)\sin \left(25\right)}{2}\end{array}}$$$$\mathrm{Green}△:$$$$\tan \left(25\right)=\frac{d}{c}=\frac{d}{1}\Rightarrow d=\tan \left(25\right)$$$$▴=\frac{dc}{2}=\frac{\tan \left(25\right)\left(1\right)}{2}=\frac{\tan \left(25\right)}{2}$$$$\mathrm{Green}▴=\frac{dc}{2}=\frac{\tan \left(25\right)}{2}$$$$\overline{)\begin{array}{c}\mathrm{Green}▴=\frac{dc}{2}=\frac{\tan \left(25\right)}{2}\end{array}}$$$$\mathrm{Red}△:$$$$anglebetweenh\mathrm{\&}g=90-2\left(25\right)=40$$$$\cos \left(40\right)=\frac{h}{g}=\frac{h}{\sqrt{c^2+d^2}}=\frac{h}{\sqrt{1+{\tan }^2\left(25\right)}}$$$$\Rightarrow h=\sqrt{1+{\tan }^2\left(25\right)}\left(\cos \left(40\right)\right)$$$$\sin \left(40\right)=\frac{i}{g}=\frac{i}{\sqrt{c^2+d^2}}=\frac{i}{\sqrt{1+{\tan }^2\left(25\right)}}$$$$\Rightarrow i=\sqrt{1+{\tan }^2\left(25\right)}\left(\sin \left(40\right)\right)$$$$▴=\frac{hi}{2}=\frac{\sqrt{1+{\tan }^2\left(25\right)}\left(\cos \left(40\right)\right)\sqrt{1+{\tan }^2\left(25\right)}\left(\sin \left(40\right)\right)}{2}$$$$\mathrm{Red}▴=\frac{hi}{2}=\frac{\left(\cos \left(40\right)\right)\left(\sin \left(40\right)\right)(1+{\tan }^2\left(25\right))}{2}$$$$\overline{)\begin{array}{c}\mathrm{Red}▴=\frac{hi}{2}=\frac{\left(\cos \left(40\right)\right)\left(\sin \left(40\right)\right)(1+{\tan }^2\left(25\right))}{2}\end{array}}$$$$$$$$\mathrm{Green}▴:\mathrm{Blue}▴:\mathrm{Red}▴$$$$=\frac{\tan \left(25\right)}{2}:\frac{\cos \left(25\right)\sin \left(25\right)}{2}:\frac{\left(\cos \left(40\right)\right)\left(\sin \left(40\right)\right)(1+{\tan }^2\left(25\right))}{2}$$$$=\tan \left(25\right):\cos \left(25\right)\sin \left(25\right):\left(\cos \left(40\right)\right)\left(\sin \left(40\right)\right)(1+{\tan }^2\left(25\right))$$

Commented by a.lgnaoui last updated on 13/Jan/24

$$\mathrm{Merci}$$

Commented by a.lgnaoui last updated on 13/Jan/24

$$\mathrm{deduire}\mathrm{the}\mathrm{Area}\mathrm{of}\mathrm{smal}\mathrm{white}$$$$\mathrm{triangle}?$$

Commented by Rasheed.Sindhi last updated on 14/Jan/24

$$Areaofwhietraingle$$$$\cos \left(25\right)=\frac{e}{d}\Rightarrow e=d\cos \left(25\right)$$$$\sin \left(25\right)=\frac{f}{d}\Rightarrow f=d\sin \left(25\right)$$$$▴=\frac{ef}{2}=\frac{d^2\cos \left(25\right)\sin \left(25\right)}{2}$$$$d=\tan \left(25\right)$$$$=\frac{{\left(\tan \left(25\right)\right)}^2\cos \left(25\right)\sin \left(25\right)}{2}$$$$$$

Commented by MathematicalUser2357 last updated on 20/Jan/24

$$\tan 25°:\frac{1}{2}\sin 130°:\frac{1}{2}\sin 100°{\sec }^{2}25°$$$$\approx 0.46:0.38:0.59$$

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