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
a) Tính các góc của tam giác
Ta có: $BC^2 = 14^2 = 196$
$AB^2 + AC^2 = 4,5^2 + 13^2 = 20,25 + 169 = 189,25$
Vì: $BC^2 > AB^2 + AC^2$ nên $\widehat{A}$ là góc tù.
Áp dụng định lý cosin:
$\cos A = \dfrac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}$
$= \dfrac{20,25 + 169 - 196}{2 \cdot 4,5 \cdot 13}$
$= \dfrac{-6,75}{117}$
$\approx -0,0577$
Suy ra: $\widehat{A} \approx 93^\circ$
Áp dụng định lý cosin với góc $B$:
$\cos B = \dfrac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$
$= \dfrac{20,25 + 196 -169}{2 \cdot 4,5 \cdot 14}$
$= \dfrac{47,25}{126}$
$\approx 0,375$
=> $\widehat{B} \approx 68^\circ$
Góc còn lại: $\widehat{C} = 180^\circ - 93^\circ - 68^\circ \approx 19^\circ$
Vậy: $\widehat{A} \approx 93^\circ,\ \widehat{B} \approx 68^\circ,\ \widehat{C} \approx 19^\circ$
b) Tính diện tích tam giác $ABC$
Nửa chu vi là: $p = \dfrac{4,5 + 14 + 13}{2} = 15,75$
Diện tích tam giác: $S = \sqrt{p(p-AB)(p-BC)(p-AC)}$
$= \sqrt{15,75(15,75-4,5)(15,75-14)(15,75-13)}$
$= \sqrt{15,75 \cdot 11,25 \cdot 1,75 \cdot 2,75}$
$\approx \sqrt{852,54}$
$\approx 29,2$
Vậy diện tích tam giác $ABC$ xấp xỉ: $29,2\ \text{đơn vị diện tích}$.
Gọi ( O;R ) , ( I ;r ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF};\widehat{BAC}=\widehat{EDF}\)) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ACB},\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)( hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OB\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
\(ID=IE\left(=r\right)\Rightarrow\Delta IDE\)cân tại I
Do đó Tam giác OAB ~ Tam giác IDE \(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\) ( đpcm)
Gọi ( O; R ), ( I; R ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF;}\widehat{BAC}=\widehat{EDF}\) ) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ABC}=\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)(hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OA\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
Do đó Tam giác OAB ~ Tam giác IDE\(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\left(đpcm\right)\)
Rất vui vì giúp đc bạn <3
BC). Trên tia đối của tia HA lấy điểm D sao cho HA = HD.Cho tam giác ABC có 3 góc đều nhọn, kẻ AH vuông góc với BC (H
