Cho ΔABC. AB = 2, AC = 5, ∠A = 120°. a) Tính BC, b) Tính SΔABC, c) Tính R, d) Tính r, e) Tính d(B; AC)

To solve the problem for triangle \( \Delta ABC \) with given values \( AB = 2 \), \( AC = 5 \), and \( \angle A = 120^\circ \), we can use the Law of Cosines and some properties of triangles to find the necessary values.

### a) Calculate \( BC \)

Using the Law of Cosines:
\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(A)
\]
\[
BC^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cdot \cos(120^\circ)
\]
Since \( \cos(120^\circ) = -\frac{1}{2} \),
\[
BC^2 = 4 + 25 + 2 \cdot 2 \cdot 5 \cdot \frac{1}{2}
\]
\[
BC^2 = 4 + 25 + 10 = 39
\]
\[
BC = \sqrt{39}
\]

### b) Calculate the area \( S_{\Delta ABC} \)

Using the formula for the area of a triangle with two sides and the included angle:
\[
S_{\Delta ABC} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(A)
\]
\[
= \frac{1}{2} \cdot 2 \cdot 5 \cdot \sin(120^\circ)
\]
Since \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \),
\[
S_{\Delta ABC} = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}
\]

### c) Calculate the circumradius \( R \)

Using the formula:
\[
R = \frac{abc}{4S}
\]
Where \( a = BC, b = CA, c = AB \). First, we calculate
\[
R = \frac{\sqrt{39} \cdot 5 \cdot 2}{4 \cdot \frac{5\sqrt{3}}{2}} = \frac{10\sqrt{39}}{10\sqrt{3}} = \frac{\sqrt{39}}{\sqrt{3}} = \sqrt{13}
\]

### d) Calculate the inradius \( r \)

Using the formula:
\[
r = \frac{S}{s}
\]
where \( s = \frac{a + b + c}{2} \).

First, we need \( s \):
\[
s = \frac{\sqrt{39} + 2 + 5}{2} = \frac{\sqrt{39} + 7}{2}
\]

Then,
\[
r = \frac{\frac{5\sqrt{3}}{2}}{\frac{\sqrt{39} + 7}{2}} = \frac{5\sqrt{3}}{\sqrt{39} + 7}
\]

### e) Calculate the distance \( d(B; AC) \)

The distance from point \( B \) to line \( AC \) can be calculated using the area:
\[
d(B; AC) = \frac{2S}{AC} = \frac{2 \cdot \frac{5\sqrt{3}}{2}}{5} = \sqrt{3}
\]

### Summary of Results

- a) \( BC = \sqrt{39} \)
- b) \( S_{\Delta ABC} = \frac{5\sqrt{3}}{2} \)
- c) \( R = \sqrt{13} \)
- d) \( r = \frac{5\sqrt{3}}{\sqrt{39} + 7} \)
- e) \( d(B; AC) = \sqrt{3} \)