Question

Problema 5, urgentttt​

Answer 1

Răspuns:

Notam cu M piciorul inaltimii AM, M ∈ BC, AM ⊥ BC

a) Aplicam teorema cosinusului pentru a afla lungimea laturii BC:

$BC=\sqrt{AB^{2} +AC^{2} -2*AB*AC*cos45}$

$BC=\sqrt{12^{2}+12^{2} -2*12*12*\frac{\sqrt{2} }{2} }$

$BC=\sqrt{12^{2}*(1+1-2*\frac{\sqrt{2} }{2} ) \\}$

$BC=12\sqrt{2-\sqrt{2} }$

$BM=6\sqrt{2-\sqrt{2} }$

În Δ AMB dreptunghic, folosim Pitagora si aflam lungimea lui AM:

$AM=\sqrt{AB^{2}-BM^{2} }$

$AM=\sqrt{12^{2} -(6\sqrt{2-\sqrt{2} } )^{2} }$

$AM=\sqrt{144-36(2-\sqrt{2}) }$

$AM=\sqrt{144-72+36\sqrt{2} }$

$AM=\sqrt{36(2+\sqrt{2} )}$

$AM=6\sqrt{2+\sqrt{2} }$

Calculăm aria Δ ABC

$A_{ABC} = \frac{AM*BC}{2}$

$A_{ABC} = \frac{6\sqrt{2+\sqrt{2} } *12\sqrt{2-\sqrt{2} } }{2}$

$A_{ABC} = 3\sqrt{2+\sqrt{2} } *12\sqrt{2-\sqrt{2} }$

$A_{ABC} = 36\sqrt{(2+\sqrt{2})(2-\sqrt{2} )}$

$A_{ABC} = 36\sqrt{4-2}$

$A_{ABC} = 36\sqrt{2} }$

b) tg(∡ABC) = AM/BM

tg(∡ABC) = $\frac{6\sqrt{2+\sqrt{2} } }{6\sqrt{2-\sqrt{2} } }=\sqrt{\frac{2+\sqrt{2} }{2-\sqrt{2} } }=\sqrt{\frac{(2+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}} =\sqrt{ \frac{6+4\sqrt{2} }{4-2} }=\sqrt{3+2\sqrt{2} }$

Explicație: