## Math 9 Chapter 1 Lesson 2: Trigonometric ratios of acute angles

## 1. Summary of theory

### 1.1. Concept of trigonometric ratio of an acute angle

**Comment: **From the above definition, it is easy to see that the trigonometric ratios of an acute angle are always positive. Furthermore we have: \(\sin\alpha < 1, \cos\alpha < 1\)

**Attention: **If two acute angles \(\alpha\) and \(\beta\) have \(\sin\alpha =\sin\beta\) ( or \(\cos\alpha =\cos\beta , \tan\alpha = ) \tan\beta ,\cot \alpha =\cot \beta\) ) then \(\alpha =\beta\) because they are corresponding angles of two similar right triangles.

### 1.2. Trigonometric ratio of two complementary angles

**Theorem: **If two angles are complementary, then the sine of one angle is equal to the cosine of the other angle, tangent of this angle is equal to the cotangent of the other angle.

Specifically in the image above with \(\alpha\) and \(\beta\) are two complementary angles, so: \(\sin\alpha =\cos\beta , \cos\alpha =\sin\beta, \tan \alpha =\cot\beta , \cot\alpha =\tan\beta\)

**Attention: **From now on, when writing trigonometric ratios of an acute angle in a triangle, we omit the “^” symbol. For example write \(\sin A\) instead of writing \(\sin\widehat{A}\)

From the definition of the trigonometric ratios of an acute angle we have: \(\tan\alpha =\frac{\sin\alpha }{\cos\alpha }; \cot\alpha =\frac{\cos\alpha } {\sin\alpha }\)

and \(\tan\alpha .\cot\alpha =1 , \sin^2\alpha +\cos^2\alpha =1\); \(1+\tan^2\alpha =\frac{1}{\cos^2\alpha }; 1+cot^2\alpha =\frac{1}{sin^2\alpha }\)

(the above formulas can be easily proved)

## 2. Illustrated exercise

### 2.1. Form 1: Calculating trigonometric ratios using the concept

**Question 1. **Let ABC be a right triangle at A, with AB=6, BC=10. Calculate sinB and cosB

**Solution guide**

We have: \(\cos B=\frac{AB}{BC}=\frac{6}{10}=0.6 ;AC=\sqrt{BC^2-AB^2}=8 \Rightarrow \sin B= \frac{AC}{BC}=0.8\)

### 2.2. Form 2: Calculating trigonometric ratios using the relationship formula between two complementary angles

**Question 1. **Convert the following trigonometric ratios into trigonometric ratios for angles less than \(45^{\circ}\) : \(\sin72^{\circ};\cos50^{\circ}; \tan68^ {\circ}; \cot 88^{\circ}\)

**Solution guide**

We have: \(\sin72^{\circ}=cos18^{\circ};\cos50^{\circ}=\sin40^{\circ}; \tan68^{\circ}=\cot22^{\circ }; \cot88^{\circ}=\tan2^{\circ}\)

**Verse 2. **Let ABC be a triangle. Knowing cosB=0.6. Calculate the trigonometric ratios of angle C.

**Solution guide**

We have: \(\sin C=\cos B=0.6\) and \(\cos C=sin B=\sqrt{1-\cos^2B}=0.8\)

\(\tan C=\frac{\sin C}{\cos C}=\frac{0.6}{0.8}=\frac{3}{4}\) and \(\cot C=\frac{\cos C}{\sin C}=\frac{0.8}{0.6}=\frac{4}{3}\)

**Verse 3.**

a) Simplify the expression: \(S=\cos^2\alpha +\tan^2\alpha .\cos^2\alpha\)

b) prove: \(\frac{(\sin\alpha +\cos\alpha )^2-(\sin\alpha -\cos\alpha )^2}{\sin\alpha .\cos\alpha }= 4\)

**Solution guide**

a) \(S=\cos^2\alpha +\tan^2\alpha .\cos^2\alpha\)

\(=\cos^2\alpha+\frac{\sin^2\alpha }{\cos^2\alpha }.\cos^2\alpha \)

\(=\sin^2\alpha +\cos^2\alpha =1\)

b) \(VT=\frac{(1+2.\sin\alpha .\cos\alpha )-(1-2.\sin\alpha .\cos\alpha )}{\sin\alpha.\cos\ alpha }=\frac{4.\sin\alpha .\cos\alpha }{\sin\alpha .\cos\alpha }=4\)

( Apply: \(\sin^2\alpha +\cos^2\alpha =1\) )

**Verse 4.** Let ABC be an acute triangle. Let a, b, c be the lengths of the sides opposite the vertices A, B, C respectively. Prove: \(\frac{a}{\sin A}=\frac{b}{\sin B} =\frac{c}{\sin C}\)

**Solution guide**

Draw AH perpendicular to BC ( \(H\in BC\) )

Then: \(\sin B=\frac{AH}{c}\Rightarrow \sin Bc=AH\) and \(\sin C=\frac{AH}{b}\Rightarrow \sin Cb=AH\)

from there we have: \(\sin Bc=\sin Cb\Rightarrow \frac{b}{\sin B}=\frac{c}{\sin C}\) .

Similarly, draw altitude BD ( \(D\in AC\) ) will prove: \(\frac{a}{\sin A}=\frac{b}{\sin B} \Rightarrow \frac{a }{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1.** Let ABC be a right triangle at A, with AC=3, BC=5. Calculate sinC and cosC.

**Verse 2.** Convert the following trigonometric ratios into trigonometric ratios of angles less than \({{45}^{{}^\circ }}\) : \(\sin {{83}^{{}^\ circ }};\,\,\,\cos {{49}^{{}^\circ }};\,\,\,\tan {{79}^{{}^\circ }};\, \,\cot {{98}^{{}^\circ }}\)

**Verse 3.** Let ABC be a right triangle at A. SinC=0.6 is known. Calculate the trigonometric ratios of angle B.

**Verse 4.** Prove the following equations:

a) \({{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}} x\)

b) \({{\sin }^{4}}x+{{\cos }^{2}}x. {{\sin }^{2}}x+{{\sin }^{2}}x= 2{{\sin }^{2}}x\)

c) \(\left( 1+\tan x \right)\left( 1+\cot x \right)-2=\frac{1}{\sin .\cos x}\)

**Question 5.** Let ABC be a right triangle at A, BC = a, CA = b, AB = c. Prove that: \(\tan \frac{B}{2}=\frac{b}{c+a}.\)

### 3.2. Multiple choice exercises

**Question 1. **The triangle OPQ has OP = 7.2, OQ = 9.6, PQ = 12. Find the measures of the angles of the triangle

A. angle O = 60, P = 50, Q = 70

B. angle O = 70, P = 50, Q = 60

C. angle O = 90, P = 53, Q = 37

D. Another result

**Verse 2. **Triangle ABC has B=60 degrees, C=45 degrees and AB=10. The perimeter of triangle ABC is

A. 35.9

B. 38.1

C. 42.5

D. 48.3

**Verse 3. **Triangle ABC, right-angled at A, has cosB = 0.8. So cotC is:

A. \(5 \over 3\)

B. \(3 \over 4\)

C. \(3 \over 5\)

D. \(3 \over 4\)

**Verse 4. **Let ABC be a right triangle at C. Know \(\cos A=\frac{5}{13}\). Then tan B=?

A. \(\frac{12}{13}\)

B. \(\frac{5}{12}\)

C. \(\frac{12}{5}\)

D. \(\frac{13}{12}\)

**Question 5. **Given acute angle \(\alpha\) know that: \(\cos\alpha -\sin\alpha =\frac{1}{5}\) The value of \(\tan\alpha\) is:

A. 1

B. \(\frac{1}{2}\)

C. \(\frac{4}{5}\)

D. \(\frac{3}{4}\)

## 4. Conclusion

Through this lesson, you should know the following:

- Concept of trigonometric ratio of an acute angle.
- The ratio of the trigonometric ratios of two complementary angles.

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