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In triunghiul ABC ascutit unghic, A=45°, AB=a si AC=[(2√2)/3]a . calculati tg B.

Răspuns :

Matematician1234
Ducem CD⊥AB.
in ΔADC: sin (∡CAD)=CD/AC=CD/2√3a/3=3CD/2√3a=3CD√3/6a=CD√3/2a=√2/2⇒ CD√3=2a√2/2=a√2 ⇒ CD=a√6 / 3 =AD (ptr ca ΔADC dreptunghic isoscel)
BD=AB-AD= a - a√6/3 = 3a-a√6 / 3 = a(3-√6) /3
in ΔBDC : tg( B) = CD/BD = a√6/3 : a(3-√6) / 3 = a√6/3 · 3/3a-a√6 = a√6 / a(3-√6) = √6 /3-√6 
R final: tg(∡B)= √6/3-√6

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