[Risolto] Trigonometria

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Ciascun lato obliquo del triangolo isoscele:
$\small  l= \dfrac{b}{2}·\cos\left[\arcsin\left(\dfrac{1}{5}\right)\right]^{-1}$

$\small l= \dfrac{\cancel{40}^{20}}{\cancel2_1}·\cos\left[11,537°\right]^{-1}$

$\small l=20·1,02062 $

$\small l= 20,4124\,u;$

altezza :

$\small h= 20·\tan(11,537°)$

$\small h= 4,0825\,u;$

perimetro $\small 2p= b+2l = 40+2×20,4124 = 80,825\,u;$

area $\small A= \dfrac{b·h}{2} = \dfrac{\cancel{40}^{20}×4,0825}{\cancel2_1} = 20×4,0825 = 81,65\,u^2.$

sin B = sin C = 1/5

angolo B = angolo C = arcsin 0,20 = 11,5370°

 tan B = 0,2041 = h/(b/2)

altezza h = 0,2041*20 = 4,0825 cm 

lato ℓ = √(b/2)^2+h^2 = √400+(50/3) = 20,412 cm

perimetro 2p = 40,824+40 = 80,824 cm

area A = 4,0825*20 = 81,650 cm^2 

Posto s = sin a = 1/5   e b = 40 

L cos a = b/2 

L = b/(2 cos a) 

P = b + 2 L = b + 2b/(2 cos a) = b( 1 + 1/sqrt(1 - s^2) ) 

h = L sin a = b s/(2 sqrt(1 - s^2)) 

S = b h/2 = b^2 s/(4 sqrt(1 - s^2)) 

Essendo 

1/sqrt(1 - s^2) = 1/sqrt(1 - 1/25) = 1/sqrt(24/25) = 5/(2 sqrt(6)) 

P = 40 ( 1 + 5/(2 sqrt(6)) = 40 + 100/sqrt(6) = 40 + 50/3 sqrt(6) = 

= 10/3 * (12 + 5 sqrt(6) ) 

numericamente 80.825 

S = 40^2/4 * 1/5 * 5/(2 sqrt(6)) = 200 sqrt(6)/6 = 100/3 * sqrt(6) 

numericamente 81.650