In △ ABC, a = 2, B = 2. If the triangle has a solution, what is the value range of a? A belongs to (0 degree, 90 degree). Why not?

In △ ABC, a = 2, B = 2. If the triangle has a solution, what is the value range of a? A belongs to (0 degree, 90 degree). Why not?

By sine theorem
a/sinA=b/sinB
sinA=asinB/b=2sinB/2√2=(√2sinB)/2
Because SINB ∈ (0.1]
Therefore, sina ∈ (0, √ 2 / 2]
And because a

In the triangle ABC, B = 2 * root sign 2A = 2 and the triangle has a solution. What is the value range of angle a?

CB=a=2,
CA=b=2√2
Draw a circle with C as the center and 2 as the radius. Point B can only be on the circle
When AB is tangent to circle C, a is the largest, and CB ⊥ AB, Sina = A / b = √ 2 / 2, a is 45 degrees
The value range of a is greater than 0 and less than or equal to 45 degrees

In △ ABC, if B = 2 2, a = 2, and the triangle has a solution, then the value range of a is () A. 0°＜A＜30° B. 0°＜A≤45° C. 0°＜A＜90° D. 30°＜A＜60°

In △ ABC, a is an acute angle. From cosine theorem, 4 = 8 + c2-4 can be obtained
2C × cosa, i.e. c2-4
2C × cosa + 4 = 0 is solvable,
The discriminant △ = 32cos2a-16 ≥ 0,  Cosa ≥
Two
2，∴0＜A≤45°，
Therefore, B

In RT △ ABC, we know that ∠ a = 60 ° and the area of △ ABC s = 12 times the root sign 3. Find a, B, C and ∠ B

∠A=60°,∠B=30°,S=ab/2 = 12√3 ,a = √3b
∴ a= 6√2,b=2√6,c=4√6

In RT △ ABC, the opposite sides of ∠ C = 90 degrees, ∠ a, ∠ B and ∠ C are respectively a, B and C. given a + B = radical 12, C = 2, calculate the area of the triangle

The square of (a + b) = 12 = the square of a + the square of B + 2A * B, because ∠ C = 90 degrees, so the square of a + the square of B = the square of C = 4, so 2A * b = 12-4 = 8, the area of triangle = 0.5 * a * b = 2

It is known that in RT △ ABC, the opposite sides of ∠ C = 90 ° and ∠ a ∠ B ∠ C are a, B, C, a + B = radical 3 + 1, C = 2, and calculate the area of △ ABC

Because (a + B ^ 2 = a ^ 2 + B ^ 2 + 2Ab = (√ 3 + 1) ^ 2 = 4 + 2 √ 3
According to Pythagorean theorem: A ^ 2 + B ^ 2 = 2 ^ 2 = 4
So 4 + 2Ab = 4 + 2 √ 3
ab=√3
Area of △ ABC = 1 / 2Ab = √ 3 / 2

It is known that in RT △ ABC, ∠ C = 90 °, the opposite sides of ∠ a, ∠ B, ∠ C are a, B and C respectively, and a + B = 2 3, C = 3, find the area of △ ABC

According to Pythagorean theorem, A2 + B2 = C2,
∵a+b=2
3，c=3，
∴（a+b）2=a2+2ab+b2=12，
∴9+2ab=12，
AB = 3
2，
The area of △ ABC = 1
2ab=1
2×3
2=3
4．

If BD: CD = 1:2, AE: CE = 3:1 in triangle ABC, the area ratio of triangle AED to triangle ABC is () Methods

If BD: CD = 1:2, AE: CE = 3:1, then the area ratio of triangle AED to triangle ABC is (3:8). Because BD: CD = 1:2, the area of triangle ACD is equal to half of the area of triangle ABC; because AE: CE = 3:1, AE = 3 / 4ac, then the area of triangle AED = 3 / 4 of the area of triangle ACD = 3 / 4 = 3

As shown in the figure, in △ ABC, ab = AC, point D is a point in △ ABC, point E and point D are on the opposite side of AC, and ad = AE, ∠ AED = ∠ ACB, then BD = CE? Please state the reasons

BD = CE, the reason is as follows: ∵ AB = AC, ad = AE,  ABC and △ ade are isosceles triangles,  ACB = ∠ ABC, ∠ AED = ∠ ade, ? AED = ∠ ACB,  ACB = ∠ ABC = ∠ AED = ∠ ade, ? ABC + ∠ ACB + ∠ BAC = 180 °, ADE + ∠ AED + ∠ DAE = 180 °, BAC = ∠ DAE

It is known that in the triangle ABC, the midpoint D and E are points on AB and AC respectively, and ad = 3 AE = 6 BD = 15 de = 5 CE = 3 BC = 15 find the angle B = angle AED

Firstly, the angle B is equal to the angle AED, and the discrimination method is as follows: in the triangle ABC and triangle AED, the three sides correspond in proportion, that is, AB / AE = BC / ed = AC / ad = 3, so there is a triangle ABC similar to the triangle AED, ad side corresponds to angle AED, AC side corresponds to angle B, so there is angle AED = angle B
By the way, remind the landlord that in the given condition be = 5, it should be de = 5, otherwise it does not hold. EC + be is greater than BC, so be is greater than 12