In ABC, it is known that b 2, a 75, b 45 , and find the edge c

1. Sinusoidal theorem: a sina = b sinb = c sinc;

2. B sin45 = C sin(180-45-75) = C sin603, B root number 2 = C root number 3;

4. c = root number 6

There is a question that c=60°, and c sinc=b sinb gives c=bsinc sinbSo c = root number 6

Over C as CD perpendicular to AB

Because the angle b is 45°, cd=db

From the right-angled triangle ACD, we can see that AD=2 cos75° CD=2 sin75°

So c=ab=ad+db=ad+cd=2(cos75°+sin75°).

The high school method of the sinusoidal theorem is obtained because a + b + c = 180 degrees, and because a b is known, c=60° is obtained because a sina = b sinb = c sinc

So 2 sin45° = c sin60

So c = root number 6

In addition, you can use the drawing method, the junior high school method, and the division method. Divide into several small triangles.

Because c=90°, a=33°15, so b and a are congruent, i.e., sinb=cosa is known by the sinusoidal theorem, a sina=b sinb=c sinc, a sina=b cosa, substitute the conditions in the problem, there is a find, b=

1、a=b=2

2. s abc=2 3 silver scramble 3

Process: (1) Feng Hedan is obtained by the sine theorem: s = 1 2 * ab * sinc beat s=1 2 * ab * 3 2 = 3ab = 4 [1].

From the cosine theorem:

c^2=a^+b^2-2ab cosc=a^2+b^2-abc^2=(a-b)^2+ab

a-b) 2=c 2-ab=2 2-4=0 [2] from [1][2] yield: a=b=2

2) From the sinusoidal theorem: a sina = b sinbsinb = 2sina

b=2a [3]

From the cosine theorem:

c 2 = a 2 + b 2-2ab cosc = a 2 + b 2-ab [3] substituting [3] to obtain: c 2 = a 2 + 4a 2-2a 2 = 3a 2 i.e. 2 2 = 3a 2 = 4

a^2=4/3

ab=a*2a=2a 2=2*4 3=8 3 is obtained by the sinusoidal theorem: s=1 2*ab*sinc

s=1 2 * 8 3 * 3 2 =2 3 3 The idea is correct, and the process should be correct, it is recommended that it is best to check it, I hope it helps

From the sinusoidal theorem, we know that a sina = b sinb

So sina = 3 2

So a = 60° or 120°

So c = 75° or 15°

So c = ( 6 + 2) 2 or ( 6 - 2) 2

Since c=90°, a=30°, we can see that abc is a right triangle, so b=180°—90°—30°=60°, because a=30° and a=3

And because c=90° b=60°, i.e. c>b

So to sum up, there are two equations:

1) The square of a = c + b = 9

2) The square of a = the square of b + the square of c.

Solve the equation c = 5, b = 4

So b=4

The hypotenuse of a 30-degree angle pair is equal to half of the hypotenuse, so a = 1 2c, then c = 6, by the Pythagorean theorem, a square + b square = c square, so that b square is equal to 27, then b = root number 27 or 3 times the root number 3The second question, c = 10, then the hypotenuse of the 30 degree angle pair is equal to half of the hypotenuse, a = 5, by the Pythagorean theorem, a square + b square = c square, b square is equal to 75, b = root number 75 or 5 times the root number 3According to the area formula, half by 5 times by 5 times the root number 5, and the calculation is done.

According to the title, the angle a is equal to 30°, and the side opposite by 30° is equal to half of the hypotenuse, i.e. c = 2b

Two straight inwards according to the right triangle.

The sum of the squares of the corners is equal to the squares of the hypotenuses, and there is.

2b）²=b²+12

3b²=12

b=2c=4

Damn, you ask this question, have you not learned acute trigonometric functions at all?

b=a/tana

c=a/sina

b = a cota = root 12 [root of 3] = 2

c=a sina = root 12 divided by [root of 2 out of 3] = 4

Because a=60°, b=30°tan, b=root, 3, so b=2, so c=4

Because c=90°, a=33°15, so b and a are redundant, i.e., sinb=cosa

From the sinusoidal theorem, a sina = b sinb sinc, a sina = b cosa, substitute the conditions in the problem, there is a find, b =

2a=3b

A = using the Pythagorean theorem a 2 + b 2 = c 2

b^2=16

b=4a=c²=52

c = 2 times the root number 13

According to the Pythagorean theorem, the square of c = the square of a + the square of b = 9 4b square + b square = 52

The solution is b=4, a=6, c=2, and 13 under the root number