## Math 10 Chapter 2 Lesson 3: Quantitative systems in triangles and solving triangles

## 1.Theory summary

### 1.1. The law of cosine in a triangle

**Theorem: **In any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides times the cosine of the angle included between them.

We have the following systems:

\(\eqalign{

& {a^2} = {b^2} + {c^2} – 2bc. {\mathop{\rm cosA}\nolimits} (1) \cr

& {b^2} = {a^2} + {c^2} – 2ac{\mathop{\rm cosB}\nolimits} (1) \cr

& {c^2} = {a^2} + {b^2} – 2ab\cos C(3) \cr} \)

\(\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}\)

\(\cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}\)

### 1.2. The sine theorem in a triangle

**Theorem: **In any triangle \(ABC\), the ratio of one side to the sine of the angle opposite that side is equal to the diameter of the circumcircle of the triangle, i.e.

\(\frac{a}{sin A}= \frac{b}{sin B} = \frac{c}{sin C} = 2R\)

where \(R\) is the radius of the circumcircle of the triangle

### 1.3. The sum of the squares of the sides and the length of the medians of the triangle

Let ABC be a triangle with median AM.

Let \(m_a;m_b;m_c\) be the medians corresponding to the sides a, b, c respectively. Then:

\(m_{a}^{2}=\frac{b^2+c^2}{2}-\frac{a^2}{4}\)

\(m_{b}^{2}=\frac{a^2+c^2}{2}-\frac{b^2}{4}\)

\(m_{c}^{2}=\frac{a^2+b^2}{2}-\frac{c^2}{4}\)

### 1.4. Triangle area

I sign h_{a}, H_{b} and h_{c} are the altitudes of the triangle \(ABC\) drawn from the families \(A, B, C\) and \(S\) respectively, which are the area of the triangle.

The area \(S\) of triangle \(ABC\) is calculated according to one of the following formulas

\(S=\frac{1}{2}a.h_a=\frac{1}{2}b.h_b=\frac{1}{2}c.h_c\)

\(S=\frac{1}{2}ab.sinC=\frac{1}{2}ac.sinB=\frac{1}{2}bc.sinA\)

\(S=\frac{abc}{4R}\)

\(S=pr\)

\(S=\sqrt{p(pa)(pb)(pc)}\)

### 1.5. Solving triangles and practical applications

Solving a triangle is calculating the lengths of the sides and the measures of the angles of a triangle based on a given condition.

To solve the triangle, we need to find the relationship between the given factors and the unknown elements of the triangle through the relations mentioned in the cosine theorem, the sine theorem and the formulas for calculating the area of the triangle. .

**Triangulation problems:** There are 3 basic problems about solving triangles:

a) Solve a triangle when one side and two angles are known.

For this problem, we use the sine theorem to calculate the remaining edge

b) Solve a triangle when knowing two sides and the included angle

For this problem we use the cosine theorem to calculate the third side

c) Solve a triangle when three sides are known

For this problem we use the cosine theorem to calculate the angle .

\(\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}\)

\(\cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}\)

**Attention: **

- It should be noted that a triangle is solvable when we know its 3 elements, of which there must be at least one length element (ie the angle factor must not be more than 2).
- Triangulation is used in practical problems, especially measurement problems.

## 2. Illustrated exercise

**Question 1: **Let ABC be a triangle with a = 7cm, b = 8cm, c = 6cm. Calculate the length of the median line m_{a} of the given triangle ABC.

**Solution guide**

\(\eqalign{

& m_a^2 = {{2({b^2} + {c^2}) – {a^2}} \over 4} = {{2({8^2} + {6^2}) – {7^2}} \over 4} = {{151} \over 4} \cr

& \Rightarrow {m_a} = {{\sqrt {151} } \over 2} \cr} \)

**Verse 2: **Let ABC be a right triangle at A inscribed in a circle of radius R and have BC = a, CA = b, AB = c.

Prove the relation: \({a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = 2R\)

Since triangle ABC is right-angled at A, the midpoint O of BC is the center of the circumcircle of triangle ABC BC = a = 2R

We have:

\(\eqalign{

& \sin A = \sin {90^0} = 1 = {a \over a} = {a \over {2R}} \cr

& \Rightarrow {a \over {\sin A}} = 2R \cr

& \sin B = {b \over a} = {b \over {2R}} \Rightarrow {b \over {\sin B}} = 2R \cr

& \sin C = {c \over a} = {c \over {2R}} \Rightarrow {c \over {\sin C}} = 2R \cr} \)

**Question 3: **Let ABC be an equilateral triangle with side a. Calculate the radius of the circumcircle of the triangle.

By the sine theorem we have:

\({a \over {\sin A}} = 2R \Rightarrow R = {a \over {2\sin A}}\)

ABC is an equilateral triangle, so A = 60^{0}

\(\eqalign{

& \Rightarrow \sin A = {{\sqrt 3 } \over 2} \cr

& \Rightarrow R = {a \over {2\sin A}} = {a \over {2. {{\sqrt 3 } \over 2}}} = {a \over {\sqrt 3 }} \cr} \)

**Question 4: **Based on formula (1) and the sine theorem. Prove: \(S = {{abc} \over {4R}}\)

According to Sin’s theorem, we have:

\({c \over {\sin C}} = 2R \Rightarrow \sin C = {c \over {2R}}\)

Then:

\(S = {1 \over 2}ab.\sin C = {1 \over 2}ab. {c \over {2R}} = {{abc} \over {4R}}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let ABC be a right triangle at A with \(\widehat B = {37^o}\) and side a=48cm. Calculate \(\widehat C\), side b, side c and height h_{a}

**Verse 2: **Let triangle ABC know sides a = 25cm; b = 36cm and c = 53cm. Calculate the angles \(\widehat A,\,\,\widehat B,\,\,\widehat C\)

**Question 3: **Find the area S of a triangle whose sides are 3, 4, and 5, respectively.

### 3.2. Multiple choice exercises

**Question 1: **The area of triangle ABC with side lengths 6, 8, 10 is:

A. \(20\)

B. \(24\)

C. \(36\)

D. \(48\)

**Verse 2: **Let ABC be a triangle with sides a, b, c 10, 15, 18 respectively. The length of the median \(b_m\) is equal to:

A. \(\approx 11.39\)

B. \(\approx 12.48\)

C. \(\approx 13.23\)

D. \(\approx 15.61\)

**Question 3: **Let a circle (O;3) inscribed in triangle ABC, right-angled at A. Knowing that triangle BDE is equilateral (as shown). The area of triangle ABC is:

A. \(9+9\sqrt{3}\)

B. \(18+9\sqrt{3}\)

C.\(18+18\sqrt{3}\)

D. \(27+18\sqrt{3}\)

## 4. Conclusion

Through this lesson, you should have a basic understanding of the following:

- Remember the formulas Theorem of cosines, Theorem of sine.
- Apply to solving triangles.

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