Using known angle in cosine theorem to find area In a triangle, if the angle a is 60 degrees, then the area s is? (expressed by a, B, c)

Using known angle in cosine theorem to find area In a triangle, if the angle a is 60 degrees, then the area s is? (expressed by a, B, c)

In the sine theorem, there are
S△=(1/2)*ab*sinC,
And (sinc) ^ 2 + (COSC) ^ 2 = 1, there is
sinC=√[1-(cosC)^2],
∴S△=(1/2)*ab*sinC=(1/2)*ab*√[1-(cosC)^2]=(√3/4)*ab.

Triangle sine cosine theorem In the unequal edge △ ABC, a is the largest edge and a ^ 2

a^2＜b^2+c^2,
b^2+c^2-a^2＞0
So cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC > 0
So a < 90
And a is the biggest side
So a > b, a > C
So 2A > b + C = 180 ° - A
So 3A > 180 degrees
So a > 60 degrees
So 60 ° a < 90 °

Application of triangle sine cosine theorem 1. In triangle ABC, a = 30 ° and B = 37 ° in scientific computer, sin37 ° is often taken as 3 / 5. If the opposite side of a is 10, the opposite side of B is B ≈? 2. In the triangle ABC, if a and B are two of the equations x ^ 2 - (√ 5) x + 1 = 0 and 2cos (a + b) = - 1, then C =? 3. In the triangle ABC, if SINB = 3 / 4, B = 10, then the value range of C is? 4. In the triangle ABC, 3A + 4B = 2c, 2A + 3B = 3C sinA:sinB :sinC=? 5. In the triangle ABC, given a = 14, B = 6, C = 10, find the maximum angle and sinc I'm sorry. The fourth question should be 3A + B = 2C

One
a/sinA=b/sinB
b=asinB/sinA=10*3/5*1/2=3
Two
2cos(A+B)=-2cos(180-A-B)=-2cosC=-1——cosC=0.5
cosC=(a^2+b^2-c^2)/2ab=[(a+b)^2-2ab-c^2]/2ab
a. B is the root, by the great theorem
(5-2-c^2)/2=0.5
C = root 3
3. No sweat
4.6c=9a+12b
6c=4a+9b
So 5A + 3B = 0
The idea is to eliminate one side and solve the relationship between the other two sides
Five
Large side to large angle, angle a is the largest
CosA=(36+100-196)/(120)=-0.5
Maximum angle a = 120
sinC/c=sinA/a
Sinc = 5 roots 3 / 14

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. It is proved that: A2-B2 c2=sin(A-B) sinC．

It is proved that by the cosine theorem A2 = B2 + c2-2bccosa,
B2 = A2 + c2-2accosb, (3 points)
The result shows that A2-B2 = b2-a2-2bccosa + 2accosb
c2=acosB-bcosA
C (6 points)
There is a sine theorem
c=sinA
sinC，b
c=sinB
Sinc, (9 points)
∴a2-b2
c2=sinAcosB-sinBcosA
sinC
=sin(A-B)
Sinc (12 points)

Draw a figure with 2 right angles, 2 obtuse angles and 1 acute angle

 

Explain why you can't draw a triangle that is not an obtuse angle, an acute angle, or a right triangle

As a plane figure, the sum of the inner angles of a triangle is only 180 degrees. No matter how you make it, one of the angles is an acute angle. The degree of the other two angles determines what triangle it is
Besides, there are only several kinds of angles in the plane, such as acute angle (< 90 degrees), obtuse angle (> 90 degrees, < 180 degrees), right angle, flat angle and rounded corner. In addition to the three angles you mentioned, it is impossible to draw a triangle if there is a flat angle, because in that case, it does not conform to the "sum of two sides is greater than the third side" in the theorem of three sides of triangle, because only the sum of two sides is greater than the third side, can there be such an angle
And rounded corners, that's even more impossible
For example, if you draw a triangle, first draw two sides, greater than or equal to the flat angle, try the existence of the third side. (1) there is a flat angle, the third side can't coincide with the two sides of the flat angle, so there is no third side 2. Look at the vertex of two sides (angle greater than 180 degrees), the third side should face this angle, But you will find that it can't intersect two outer edges (even if the two sides are regarded as two rays from a vertex). If it can't intersect, it can't form another two vertices, so it can't form a triangle, so here, the third side can't exist

The three heights of acute triangle are in (), obtuse triangle has () height, right triangle has two heights which are exactly its () 1. Three heights of acute triangle are in (), obtuse triangle has () height outside triangle, right triangle has two heights which are exactly its () 2. The relationship between the three sides of a triangle 3. In △ ABC, if ∠ B - ∠ a - ∠ C = 50 °, B = ()

Acute triangle has three heights (inside triangle), obtuse triangle has (2) height outside triangle, right triangle has two heights which are exactly its right side
The relationship between the three sides of a triangle (the sum of any two sides is greater than the third side, and the difference between any two sides is less than the third side)
3. In △ ABC, if ∠ B - ∠ a - ∠ C = 50 °, B = (115 °)

What are the three heights of right triangle, acute triangle and obtuse triangle

The three heights of a triangle meet at a point called the perpendicular of the triangle
The center of the acute triangle is inside the triangle, the center of the right triangle is the vertex of the right angle, and the vertical center of the obtuse triangle is outside the triangle (obtained by the intersection of high extension lines)

Each acute triangle can be divided into right triangle and obtuse triangle?

A: No
An acute triangle can be divided into two right triangles
But it can not be divided into two obtuse angle triangles
The dividing line must pass through the vertex and the opposite side, and the top angle to be divided into two smaller acute angles,
The 180 ° edge of the opposite side can only be divided into an obtuse angle and an acute angle,
Therefore, only one obtuse angle can be segmented at most, and two obtuse angle triangles cannot be separated

Judge whether △ ABC is an acute triangle, a right triangle or an obtuse triangle 1.a=5,b=12,c=13 2. Vertex coordinates a (6, - 7) B (7,5) C (2,3) 3. A = 2 b = root 2 C = root 3 + 1

1. Use Pythagorean theorem! 5 ^ 2 + 12 ^ 2 = 169 = 13 ^ 2 right triangle
2. Use vector! Vector CA = (4, - 10), vector CB = (5,2), ca * CB = 20-20 = 0 right triangle
3. Use cosine theorem! 4 + 2 - (radical 3 + 1) ^ 2 = 2-2 radical 3