Ratio-of-green-blue-and-Red-triangles-Evaluate-these-Area-for-25-and-length-of-aquart-1-

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

Ratio of green blue and Red triangles ?

Evaluate these Area for α = 25 ° and

length of 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

Blue △ :

\cos (25) = \frac{a}{c} = \frac{a}{1} \Rightarrow a = \cos (25)

\sin (25) = \frac{b}{c} = \frac{b}{1} \Rightarrow b = \sin (25)

Blue ▴ = \frac{a b}{2} = \frac{\cos (25) \sin (25)}{2}

\begin{matrix} Blue ▴ = \frac{a b}{2} = m \frac{\cos (25) \sin (25)}{2} \end{matrix}

Green △ :

\tan (25) = \frac{d}{c} = \frac{d}{1} \Rightarrow d = \tan (25)

▴ = \frac{d c}{2} = \frac{\tan (25) (1)}{2} = \frac{\tan (25)}{2}

Green ▴ = \frac{d c}{2} = \frac{\tan (25)}{2}

\begin{matrix} Green ▴ = \frac{d c}{2} = \frac{\tan (25)}{2} \end{matrix}

Red △ :

a n g l e b e t w e e n h & g = 90 - 2 (25) = 40

\cos (40) = \frac{h}{g} = \frac{h}{\sqrt{c^{2} + d^{2}}} = \frac{h}{\sqrt{1 + \tan^{2} (25)}}

\Rightarrow h = \sqrt{1 + \tan^{2} (25)} (\cos (40))

\sin (40) = \frac{i}{g} = \frac{i}{\sqrt{c^{2} + d^{2}}} = \frac{i}{\sqrt{1 + \tan^{2} (25)}}

\Rightarrow i = \sqrt{1 + \tan^{2} (25)} (\sin (40))

▴ = \frac{h i}{2} = \frac{\sqrt{1 + \tan^{2} (25)} (\cos (40)) \sqrt{1 + \tan^{2} (25)} (\sin (40))}{2}

Red ▴ = \frac{h i}{2} = \frac{(\cos (40)) (\sin (40)) (1 + \tan^{2} (25))}{2}

\begin{matrix} Red ▴ = \frac{h i}{2} = \frac{(\cos (40)) (\sin (40)) (1 + \tan^{2} (25))}{2} \end{matrix}

Green ▴ : Blue ▴ : Red ▴

= \frac{\tan (25)}{2} : \frac{\cos (25) \sin (25)}{2} : \frac{(\cos (40)) (\sin (40)) (1 + \tan^{2} (25))}{2}

= \tan (25) : \cos (25) \sin (25) : (\cos (40)) (\sin (40)) (1 + \tan^{2} (25))

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

Merci

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

deduire the Area of smal white

triangle ?

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

A r e a o f w h i e t r a i n g l e

\cos (25) = \frac{e}{d} \Rightarrow e = d \cos (25)

\sin (25) = \frac{f}{d} \Rightarrow f = d \sin (25)

▴ = \frac{e f}{2} = \frac{d^{2} \cos (25) \sin (25)}{2}

d = \tan (25)

= \frac{{(\tan (25))}^{2} \cos (25) \sin (25)}{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