problema geometria

ΑΒ = 24 cm ; CH = ? ; AK = ?

ΑΗ = ΗΒ = 12 cm

SIN(γ/2) = √((1 - COS(γ))/2) = √((1 - 7/25)/2) = 3/5

COS(γ/2) = √(1 - (3/5)^2) = 4/5

TAN(γ/2) = 3/5/(4/5) = 3/4

3/4 = 12/CH---> CH = 4·12/3-----> CH = 16 cm

l = √(12^2 + 16^2) = 20 cm lato obliquo

A = 1/2·24·16  = 192 cm^2 area triangolo

ΑΚ = 2·A/l = 2·192/20----> ΑΚ = 19.2 cm

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Angolo al vertice $\alpha= \cos^{-1}\left(\dfrac{7}{25}\right) = 73,74°;$ $^{(1)}$

ciascun angolo alla base $\beta= \dfrac{180°-73,74°}{2} = \dfrac{106,26°}{2} = 53,13°;$

altezza relativa alla base $h= \dfrac{b}{2}×\tan(\beta) = \dfrac{24}{2}×\tan(53,13°) = 16\,cm;$

ciascun lato obliquo $l= \dfrac{b}{2}×\cos(\beta)^{-1} = \dfrac{24}{2}×cos(53,13)^{-1} = 20\,cm;$

area $A= \dfrac{b×h}{2} = \dfrac{24×\cancel{16}^8}{\cancel2_1} = 24×8 = 192\,cm;$

altezza relativa al lato obliquo $h_1= \dfrac{2A}{l} =  \dfrac{\cancel2^1×192}{\cancel{20}_{10}}= \dfrac{192}{10} = 19,2\,cm.$ 

Note:

$^{(1)}$ - $ \;\cos^{-1} =$ arcoseno.

b = 24

cos  = 7/25

 = arccos 7/25 =73,74°

Â/2 = 36,87°

tan Â/2 = 0,750 = (b/2)/h

base b = 1,5h 

altezza h = 24/1,5 = 16 cm 

lato l = √h^2+(b/2)^2 = 4√4^2+3^2 = 4*5 = 20 cm 

altezza h' = b*h/l = 24*16/20 = 19,20 cm