Bài tập Hệ thức lượng trong tam giác vuông có lời giải

In the following article, we re-systematize the formulas of the quantification system in a right triangle, thereby applying these formulas to solve some types of illustrative exercises to better understand and remember the relational relations. this weight.

I. The trigonometric system in right triangles to remember

Let ABC have a square at A (angle A is 90⁰) as shown below:

Yes: AH ⊥ BC, AB = c, AC = b, BC = a, AH = h, then:

• BH = c’ is called the projection of AB onto BC

• CH = b’ is called the projection of AC down BC

Then we have:

1) AB² = BH.BC or c² = ac’

AC² = CH.BC or b² = ab’

2) AH² = CH.BH or h² = b’.c’

3) AB.AC = AH.BC or bc = ah

4) 1/(AH)² = 1/(AB)² + 1/(AC)² or 1/h² = 1/b² + 1/c²

5) AB² + AC² = BC² good² + c² = a² (Python’s Theorem)

II. Trigonometric ratio of acute angle

1. Definition:

– Let ABC be a triangle (right at A) including the opposite side, hypotenuse and adjacent side:

• sinα = Opposite/Huyen = AB/BC

• cosα = Ke/Huen = AC/BC

• tanα = Opposite/Adjacent = AB/AC

• cotα = Adjacent/Opposite = AC/AB

2. Compare trigonometric ratios

a) Let α and β be two acute angles. If α < β then

• sinα < sinβ; tanα < tanβ

• cosα > cosβ; cotα > cotβ

b) sinα < tanα; cosα < cotα

→ You can refer to the full formula in the article: Trigonometric formula in a right triangle

III. Exercises on trigonometry in right triangles

* Exercise 1: Let ABC be a right triangle at A. In which AB = 12cm, AC = 9cm. Calculate the trigonometric ratios of angle B, then deduce the trigonometric ratios of angle C.

* The answer:

According to the Pythagorean theorem we have:

So, we have:

Since angle B and angle C are complementary angles:

* Exercise 2: Let ABC be a right triangle at A, altitude AH. Know AC = 20 cm, BH = 9 cm. Calculate the lengths BC and AH

> Solution:

• We set HC = x (x>0).

Apply the AC . formula² = BC.HC, we get:

20² = (9 + x)x

x² + 9x – 400 = 0

(x + 25)(x – 16) = 0

⇔ x = -25 (type) or x = 16

So the length of the hypotenuse BC is:

BC = BH + HC = 9 + 16 = 25 cm

– I have: AH² = HB.HC = 9.25 = 3².5² = 15²

So the length of the altitude AH is: AH = 15 (cm)

* Exercise 3: Let ABC be a right triangle at A, AB : AC = 7 : 24, BC = 625 cm. Calculate the length of the projection of the two sides of the right angle on the hypotenuse.

* The answer:

We have the following illustration:

Drawing AH ⊥ BC, then we have:

AB² = BH.BC ;

AC² = CH.BC ;

So we have:

(using the proportionality property: )

Infer: BH = 49.1 = 49;

CH = 576.1 = 576

* Exercise 4: Let ABC be a right triangle at A, BC = a, CA = b, AB = c.

Prove that:

* The answer:

We draw the following figure:

From angle B draw bisector BD. Then we have:

According to the bisector properties, we have:

Consider triangle ABD, right angled at A, with:

So we have something to prove.

* Exercise 5: Prove that the values ​​of the following expressions do not depend on the values ​​of acute angles α, β

a) cos² α.cos² β + cos² a.sin² β + sin² α

b) 2(sin⁡α – cos⁡α )² – (sin⁡α + cos⁡α )² + 6sin⁡α.cos⁡α

c) (tan⁡α – cot⁡α )² – (tan⁡α + cot⁡α )²

* The answer:

a) cos² α.cos² β + cos² a.sin² β + sin² α

= cos² = cos² α(cos² β + sin² β) + sin² α

= cos² α.1 + sin² α

= 1

b) 2(sin⁡α – cos⁡α )² – (sin⁡α + cos⁡α )² + 6 sin⁡α.cos⁡α

= 2(1 – 2sinα.cos⁡α ) – (1 + 2sinα.cos⁡α ) + 6sinα.cos⁡α

= 1 – 6sinα.cos⁡α + 6sinα.cos⁡α

= 1

c) (tan⁡α – cot⁡α )² – (tan⁡α + cot⁡α )²

= (tan² α – 2 tan⁡α.cotα + cot² α) – (tan² α + 2 tan⁡α.cotα + cot² a )

= -4 tan⁡α.cotα

= -4.1 = -4

Hope through the content about The trigonometric system in the 9th grade right triangle and illustrated math exercises The above help students better remember, master, and easily apply these quantifiers in similar types of exercises. If you have any suggestions and questions, please leave a comment below the article for Hay Learn to recognize and support, wish you good luck in your studies.