Question

[Nguyễn Thái Sơn]

1) Tam giac ABC co AB = \(\dfrac{\sqrt{6}-\sqrt{2}}{2}\), BC = \(\sqrt{3}\), CA = \(\sqrt{2}\) . Goi D la chan duong phan giac trong goc A. Khi do goc ADB bang bao nhieu do

A. 45^o       B.60^o        C.75^o          D. 90^o

Answer 1

[Akai Haruma]

Lời giải:Áp dụng định lý cos ta có:

\(\cos A=\frac{AB^2+AC^2-BC^2}{2AB.AC}=\frac{-1}{2}\Rightarrow \widehat{A}=120^0\)

\(\cos B=\frac{BC^2+BA^2-AC^2}{2BC.BA}=\frac{-\sqrt{2}}{2}\Rightarrow \widehat{B}=45^0\)

\(\widehat{C}=180^0-(\widehat{A}+\widehat{B})=180^0-(120^0+45^0)=15^0\)

\(\widehat{ADB}=180^0-(\frac{\widehat{A}}{2}+\widehat{B})=180^0-(\frac{120^0}{2}+45^0)=75^0\)