tính sin 36 độ
tính sin 36 độ
Bắt đầu bởi hienhienhien, 03-05-2017 - 21:15
#1
Đã gửi 03-05-2017 - 21:15
#2
Đã gửi 03-05-2017 - 21:34
bấm máy đc mà
#3
Đã gửi 03-05-2017 - 22:17
tinh so chinh xac bang phuong phap hinh hoc
#4
Đã gửi 03-05-2017 - 22:45
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Xét bài toán sau: Cho tam giác $ABC$ cân tại $A$ có $\widehat{A}=36^\circ$, $\widehat{B}=\widehat{C}=72^\circ$, $BC=1$. Vẽ đường phân giác $CD$ và đường cao $CH$ của tam giác.
Chú ý rằng $\widehat{BCD}=\widehat{DCA}=\widehat{BAC}\quad \left(=36^\circ\right)$ nên $\triangle BCD$ cân tại $C$ và $\triangle ADC$ cân tại $D$. Vì vậy ta có $AD=DC=BC=1$.
Tam giác cân $BCD$ có đường cao $CH$, đồng thời $CH$ là đường trung tuyến. Vậy đặt $HD=HB=x$.
Xét $\triangle ABC$ và $\triangle BCD$ có
$\left.\begin{array}{l} \widehat{BAC}=\widehat{DCB} \quad \left(=36^\circ\right) \\ \widehat{B} \text{ chung} \end{array}\right \} \implies \triangle ABC \backsim \triangle CDB \quad \text{(g.g)}$
Vậy ta có
\[\dfrac{AB}{CD}=\dfrac{BC}{BD}\iff 1+2x=\dfrac{1}{2x}\iff 2x+4x^2=1 \iff \left[ \begin{array}{l} x=\dfrac{\sqrt{5}-1}{4} \\ x=-\dfrac{\sqrt{5}+1}{4}<0 \end{array} \right.\]
Khi đó, trong tam giác $AHC$ ta có $\sin 36^\circ=\sin A= \dfrac{HC}{AC}=\dfrac{\sqrt{AC^2-AH^2}}{AC}=\dfrac{\sqrt{\left(1+2x\right)^2-\left(1+x\right)^2}}{1+2x}=\sqrt{\dfrac{5-\sqrt{5}}{8}}$.
Bài viết đã được chỉnh sửa nội dung bởi huykinhcan99: 03-05-2017 - 22:46
- HoangKhanh2002 yêu thích
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