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

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

 

Is a right triangle an acute triangle or an obtuse triangle?

Triangles are divided into acute angles, right angles and obtuse angles according to angles. Right triangles are neither acute nor obtuse angles

In a triangle, a = 60 °, BC = 7, ab = 5, then the area of ABC in the triangle is? Can you solve it with the sine cosine theorem,

First make a graph, let BC = a, ab = C, AC = B
First, use the cosine theorem to find AC
2bccosA=c2+b2-a2
Substitute B = 8, B = - 3, round off
According to the sine theorem, s = bcsina / 2 = 8 * 5 * radical 3 / 4 = 10 radical 3

In triangle ABC, always a = 6, B = 7, C = 5, find the area s of triangle ABC. Do it by the method of cosine theorem,

Helen formula:
There is a triangle with side lengths a, B and C respectively. The area s of the triangle can be obtained by the following formula:
S=√[p(p-a)(p-b)(p-c)]
P in the formula is half circumference:
p=(a+b+c)/2

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)
‡ A2-B2 = b2-a2-2bccosa + 2accosb to obtain A2-B2
c2=acosB-bcosA
C (6 points)
According to the sine theorem, there is a
c=sinA
sinC，b
c=sinB
Sinc, (9 points)
∴a2-b2
c2=sinAcosB-sinBcosA
sinC
=sin(A-B)
Sinc (12 points)

Application of triangle sine and cosine theorem 1. In triangle ABC, a = 30 °, B = 37 °, in scientific computer, sin37 ° ≈ 3 / 5 is often taken. If the opposite side of a is a = 10, then the opposite side of B is B ≈? 2. In triangle ABC, if a and B are two of equation x ^ 2 - (√ 5) x + 1 = 0, and 2cos (a + b) = - 1, then C =? 3. In triangle ABC, if SINB = 3 / 4 and B = 10, the value range of C is? 4. In triangle ABC, 3A + 4B = 2c, 2A + 3B = 3C, then Sina: SINB: sinc =? 5. In triangle ABC, given a = 14, B = 6, C = 10, find the maximum angle and sinc Er... 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 = piece 3
3. Well, no sweat
4.6c=9a+12b
6c=4a+9b
So 5A + 3B = 0 - (the title is wrong.)
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 pcs 3 / 14