## Math 9 Chapter 1 Lesson 4: Some relations about sides and angles in a right triangle

## 1. Summary of theory

### 1.1. The systems

**Theorem: T**In a right triangle, each side of the right angle is equal to:

a) The hypotenuse multiplied by the sine of the opposite angle or multiplied by the cosine of the adjacent angle.

b) The side of the other right angle is multiplied by tan of the opposite angle or multiplied by the cotan of the adjacent angle.

Specifically, in the above triangle:

\(b=a.sinB=a.cosC;c=a.sinC=a.cosB\)

\(b=c.tanB=c.cotC;c=b.tanC=b.cotB\)

### 1.2. Apply the solution of right triangle

Let ABC be a right triangle at A with BC=10 and \(\widehat{C}=30^{\circ}\). Solve right triangle ABC.

**Solution guide**

We can easily deduce: \(\widehat{B}=60^{\circ}\)

\(AC=BC.cosC=10.cos30^{\circ}=10.\frac{\sqrt{3}}{2}=5.\sqrt{3}\)

\(AB=BC.sinC=10.sin30^{\circ}=10.\frac{1}{2}=5\)

## 2. Illustrated exercise

**Question 1. **Let ABC be a right triangle at A, BC=14. Calculate the length AB, AC knowing AB=AC

**Solution guide**

Since AB=AC, triangle ABC is right-angled at A, so: \(\widehat{B}=\widehat{C}=45^{\circ}\)

Then: \(AB=AC=BC.sin45^{\circ}=14.\frac{1}{\sqrt{2}}=7\sqrt{2}\)

**Verse 2. **A tree is 10m high at a time when its shadow on the ground is 4m. Calculate the angle (rounded to the minute) that the sun’s rays make with the earth.

**Solution guide: **

We consider the tree, the direction of the light ray and the ground to form a triangle ABC as shown

Then: \(\widehat{ACB}\) is the angle that the sun’s rays make with the ground

We have: \(tan\widehat{ACB}=\frac{AB}{BC}=\frac{10}{4}=\frac{5}{2}\Rightarrow \widehat{ACB}\approx 60^{ \circ}12{}’\)

**Verse 3.** A boat crosses a river with a speed of 3 km/h in 6 minutes. Know that the direction of the boat makes an angle of 60 degrees with the shore. Calculate the width (m) of that river section

**Solution guide: **

We will see the description of the problem as shown in the figure. We will solve the problem through solving triangle ABC as shown

easy to see \(\widehat{BAC}=30^{\circ}\) we will change the unit: \(3km/h=\frac{5}{6}m/s\), 6 minutes=360 (S)

\(\Rightarrow S=\frac{5}{6}.360=300(m)\)\(\Rightarrow AB=S.cos30^{\circ}=300.\frac{\sqrt{3}}{ 2}=150\sqrt{3}(m)\)

**Verse 4. **Prove that the area of a triangle is half the product of the two sides times the sine of the acute angle formed by the lines containing the two sides.

**Solution guide**

With triangle ABC, set the angle formed by AB and AC as \(\alpha\). we will Cm: \(S_{\Delta ABC}=\frac{1}{2}.AB.AC.sin\alpha\)

We have: \(\Delta AHC\) square at H so: \(CH=AC.sin\alpha\)

\(S_{\Delta ABC}=\frac{1}{2}.AB.CH=\frac{1}{2}.AB.AC.sin\alpha\)

**Question 5. **Let ABC be an acute triangle with BC=a, AB=c, AC=b. Prove that: \(a^2=b^2+c^2-2.bccosA\)

**Solution guide**

Draw CD perpendicular to AB (D belongs to AB)

Set \(AD=b’,BD=a’\Rightarrow c=a’+b’\)

Applying the Pythagorean theorem we have: \(a^2=a’^2+CD^2=(c-b’)^2+b^2-b’^2=c^2+b ^2-2.c.b’\)

where \(b’=b.cosA\Rightarrow a^2=b^2+c^2-2.bccosA\)

## 3. Practice

### 3.1. Essay exercises

**Question 1. **Let ABC be a right triangle at A, BC = 20cm. Find the length of AB and AC knowing AB=AC.

**Verse 2. **A tree is 8m tall at a time when its shadow on the ground is 6m. Calculate the angle (rounded to the minute) that the sun’s rays make with the earth.

**Verse 3. **A canoe crosses a river with a speed of 20 km/h in 2 minutes. Know that the direction of the boat makes an angle of 30 degrees with the shore. What is the width (m) of that river?

**Verse 4. **Let ABC be an acute triangle with BC=a, AB=c, AC=b. Prove that: \({{c}^{2}}={{a}^{2}}+{{b}^{2}}-2.ab\cos C\)

### 3.2. Multiple choice exercises

**Question 1. **Let ABC be a right triangle at A with BC=8, \(\widehat{C}=30^{\circ}\). \(S_{\Delta ABC}\) has a value of:

A. \(8\sqrt{2}\) B. \(8\sqrt{3}\)

C. \(4\sqrt{3}\) D. \(4\sqrt{2}\)

**Verse 2.** Let ABC be a right triangle at A with AB-AC=6. Knowing that \(\widehat{C}=\alpha, tan\alpha =3\). The value of \(S_{\Delta ABC}\) is:

A. 13 B. 14.5

C. 13.5 D. 14

**Verse 3. **Let ABCD quadrilateral with AC=8, BD=10, \(\widehat{AOB}=30^{\circ}\). The value of \(S_{ABCD}\) is:

A. 20 B. 22

C. 12 D. 18

**Verse 4. **Given 2 right triangles ABC (at A) and A’B’C’ (at A’) have: \(\widehat{C}=30^{\circ}\),BC=6, \(\ widehat{C’}=45^{\circ}\), B’C’=4. Compare AB and A’B’

A. \(AB>A’B’\)

B. \(A’B’>AB\)

C. \(AB=A’B’\)

D. Incomparable

**Question 5.** Let ABC be a right triangle at A with BC=6. \(\widehat{B}=60^{\circ}\). high road AH. Let E and F be the perpendicular projection of H onto AB, AC . respectively

What is the value of the expression S=AE.AB+AF.FC

A. \(\frac{27}{4}\)

B. \(\frac{27}{2}\)

C. \(\frac{27\sqrt{3}}{2}\)

D. \(\frac{27}{\sqrt{2}}\)

## 4. Conclusion

Through this lesson, you should know the following:

- Master the relations for sides and angles in right triangles.
- Apply knowledge to solve problems about right triangles.

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