Question

Cho ΔABC có góc A =60*, góc B=45*, b=8.

a) Tính a,c, góc C

b) Tính diện tích tam giác

 

Answer 1

a, Kẻ \(CH\perp AB\Rightarrow CH=AC.sin60^o=\dfrac{8.\sqrt{3}}{2}=4\sqrt{3}\)

\(\Rightarrow BC=\dfrac{CH}{sin45^o}=\dfrac{4\sqrt{3}}{\dfrac{\sqrt{2}}{2}}=4\sqrt{6}\)

\(AH=AC.cosA=8.cos60^o=4\)

\(BH=\dfrac{CH}{tan45^o}=4\sqrt{3}\)

\(\Rightarrow AB=AH+BH=4\sqrt{3}+4\)

\(\widehat{C}=180^o-\widehat{A}-\widehat{B}=180^o-60^o-45^o=75^o\)

b, \(S_{ABC}=\dfrac{1}{2}.AC.AB.sinA=\dfrac{1}{2}.8.\left(4+4\sqrt{3}\right).sin60^o=24+8\sqrt{3}\)

Answer 2

a) \(\widehat{C}\)= 180-(60+45)=75^o