The topics|Heim

8

THERE ARE TWOSpecial triangles in trigonometry. One is the 30°-60°-90° triangle. The other is theisosceles right triangle. They are special because with simple geometry we know the ratios of their sides and can therefore solve any such triangle.

Sentence. *In a 30°-60°-90° triangle the sides are in the ratio 1 : 2 :.*

We will prove thatunder.

Note that the smallest side 1 is opposite the smallest angle 30°; while the greatest side, 2, opposes the greatest angle, 90°. (set 6). (Because 2 is greater than. Also during 1 :: 2 correctly corresponds to opposite sides 30°-60°-90°, many find the order 1 : 2 :easier to remember.)

The quoted theorems are from the appendix,Some theorems of plane geometry.

Here are examples of how we use knowledge of these ratios. First we can evaluate the functions of 60° and 30°.

Example 1.Evaluatecos 60°.

**Answer**.For any problem involving a 30°-60°-90° triangle, the student should not use a table. The student should sketch the triangle and place the ratio numbers.

Since the cosine is the ratio ofSide to the hypotenuse, we can see that

cos 60° = ½.

example 2Evaluate sin 30°.

**Answer**.After the property ofcofunctions, sin is 30°*even*Tocos60°. sin 30° = ½.

On the other hand, you can see that directly in the image above.

Problem 1.Evaluatesin 60° and tan 60°.

To see the answer, hover over the colored area with your mouse.

To cover the answer again, click Refresh.

The sine is the ratio of the opposite side to the hypotenuse.

sin 60° = | 2 | = ½. |

Lesson 5 of Algebra

The tangent is the ratio of the opposite side to the adjacent.

hellbraun 60° = | 1 | =. |

Problem 2.EvaluateCot 30° and Cos 30°.

The cotangent is the ratio of the adjacent side to the opposite side.

Therefore, looking at the figure above, cot 30° = | 1 |

=.

Or more simply: cradle 30° = tan 60°.

Problem 1

The cosine is the ratio of the adjacent side to the hypotenuse. For this reason,

cos 30° = | 2 | = ½. |

Before we get to the next example, here's how we relate the sides and angles of a triangle:

If an angle is marked with a capital A, the opposite side is marked with a small one*A*. Similarly for angle B and side*B*, angle C and side*C*.

Example 3.Solve right triangle ABC when angle A is 60° and side AB is 10 cm.

**Solution.**TosolveA triangle means knowing all three sides and all three angles. Since this is a right triangle and angle A is 60°, the remaining angle B is its complement, 30°.

Here, too, the sides are in a ratio of 1:2 in every 30°-60°-90° triangle:, as shown on the left.

If we know the aspect ratios, we don't need the trigonometric functions or the Pythagorean theorem to solve a triangle. We can solve it through themethod of similar numbers.

Now the sides that form equal angles have the same ratio. Proportionally,

2 : 1 = 10 : alternating current.

2 is twice 1. Therefore 10 is twice AC. AC is 5cm.

The side adjacent to 60°, wesee, is always*half*the hypotenuse.

As for BC—proportionally,

2 := 10 : BC.

To get 10, 2 was multiplied by 5. thats whyWilleAlsobe multiplied by 5. BC is 5cm.

In other words, since one side of the standard triangle has been multiplied by 5, each side will be multiplied by 5.

1 : 2 := 5 : 10 : 5.

Compare example 11Here.

Again, if we know the ratios, the student should use them to solve the trianglemethod of similar numbers, not the trigonometric functions.

(InThema10, we solve right triangles whose aspect ratios we don't know.)

task 3In the right triangle DFE, the angle D is 30° and the side DF is 3 inches. How long are sides*D*And*F*?

The student should draw a similar triangle in the same orientation. Then you will see that the page matcheswas multiplied by.

Lesson 26 of Algebra

Therefore each side is multiplied by. Page*D*will be

1=. Page*F*will be 2nd.

task 4In the right triangle PQR, the angle P is 30° and the side*R*is 1cm. How long are sides*P*And*Q*?

The side corresponding to 2 was*divided*by 2. Therefore, each side must be divided by 2. Page*P*will be ½ and side*Q*will be ½.

task 5Solve right triangle ABC when angle A is 60° and hypotenuse is 18.6 cm.

The side adjacent to 60° is always*half*the hypotenuse - i.e. side*B*is 9.3 cm.

But that's the side that corresponds to 1. And it was multiplied by 9.3. Therefore page*A*is multiplied by 9.3.

It will be March 9thcm.

task 6Prove:*The area A of an equilateral triangle whose side is s is*

*A*= ¼*S*^{2}.

The area*A*of any triangle is equal to half the sine of any angle multiplied by the product of the two sides that form the angle. (Thema 2, Problem 6.)

In an equilateral triangle is every side*S*, and each angle is 60°. For this reason,

*A*= ½ sin 60°*S*^{2}.

And sin 60° = ½,

Problem 1

*A*= ½**·**½ *S*^{2}= ¼*S*^{2}.

task 7Prove:*The area A of an equilateral triangle inscribed in a circle of radius r is*

A= | 34 | R^{2}. |

The three radii divide the triangle into three congruent triangles.

side-side-side

Therefore, each radius bisects each vertex into two 30° angles.

If we increase the radius AO, then AD is the perpendicular bisector of side CB.

sentence 2

So Triangle OBD is a 30-60-90 triangle.

Let's call each side of the equilateral triangle*S*, then in the right triangle OBD,

½SR | = | cos 30°=½. |

For this reason,

*S*=*R*

so that

*S*^{2}= 3*R*^{2}.

Well, the area*A*of an equilateral triangle

*A*= ¼*S*^{2}.

task 6

For this reason,

A= ¼S^{2}= ¼·3R^{2} | = | 34 | R^{2}. |

We wanted to prove that.

task 8Prove:*The bisectors of an equilateral triangle meet at a point two-thirds the distance from the triangle's apex to the base.*

Let ABC be an equilateral triangle, let AD, BF, CE be the bisectors of angles A, B, and C, respectively; then these bisecting lines meet at point P, so that AP is two-thirds of AD.

First, the triangles BPD, APE are congruent.

For since the triangle is equilateral and BF, AD are the bisecting lines, the angles PBD, PAE are equal and each 30°;

and side BD is equal to side AE, for in an equilateral triangle the bisector is the perpendicular bisector of the base.

sentence 2

The angles PDB, AEP are then right angled and equal.

For this reason,

angle-side-angle

Triangles BPD, APE are congruent.

Now, | BPPD | =csc 30° = 2. |

Problem 2

Hence BP = 2PD.

But AP = BP, because the triangles APE, BPD coincide, and these are the sides opposite the same angles.

Therefore AP = 2PD.

Therefore AP is two thirds of the total AD.

What we wanted to prove.

Theprove

Here is proof that in a 30°-60°-90° triangle the sides are in a 1:2 ratio:. It is based on the fact that it is a 30°-60°-90° triangle*half*of an equilateral triangle.

Draw the equilateral triangle ABC. Then each of its equal angles is 60°. (sentences 3 and 9)

PullDieStraight line AD bisecting the angle at A into two 30° angles.

Then AD is the perpendicular bisector of BC (sentence 2). The triangle ABD is therefore a 30°-60°-90° triangle.

Now since BD equals DC, BD is half of BC.

This implies that BD is also half of AB since AB equals BC. That is,

BD : AB =1:2

Of thePythagorean theorem, we can find the third side AD:

ADVERTISEMENT^{2}+1^{2} | = | 2^{2} |

ADVERTISEMENT^{2} | = | 4 - 1 = 3 |

ADVERTISEMENT | = | . |

Therefore, in a 30°-60°-90° triangle, the sides are in the ratio 1 : 2 :; we wanted to prove that.

Logical order. The square drawn at the height of an equilateral triangle is three quarters of the square drawn on the side.

Next topic: The isosceles right triangle

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