It is known that in △ ABC, ∠ C = 90, ∠ a = 60, a + B = 3 + radical 3, then a is equal to

It is known that in △ ABC, ∠ C = 90, ∠ a = 60, a + B = 3 + radical 3, then a is equal to

Because: △ ABC is RT △ and ∠ a = 60 °,
Therefore: TG ∠ a = tg60 ° = √ 3 = A / B, that is, a = (√ 3) B
And a + B = 3 + √ 3
By solving the equation composed of these two equations, a = 3, B = √ 3

In △ ABC, if ∠ a = 60 °, B = 45 ° and BC = 3 2, then AC = () A. 4 Three B. 2 Three C. Three D. Three Two

According to the sine theorem, BC
sinA＝AC
sinB，
Then AC = BC · SINB
sinA＝3
2 x
Two
Two
Three
2＝2
Three
Therefore, B is selected

In △ ABC, if a = radical 3-1, B = radical 3 + 1, C = radical 10, then the degree of the maximum angle of △ ABC is

C is the largest edge, then C is the largest angle
According to the cosine theorem:
cosC=(a^2+b^2-c^2)/2ab=(4+2√3+4-2√3-10)/2(√3+1)(√3-1)
= -2/4= -1/2
C = 120 degrees
☆⌒_ I hope I can help you~

In the right triangle ABC, the angle c = 90 degrees, a > b, a + B = (root 3 + 1) / 2 * C, find a, B

A + B = (radical 3 + 1) / 2 * C
Sina + SINB = radical 3 / 2 + 1 / 2
Sina = radical 3 / 2
A=60°
B=30°
Another explanation
(a + b) ^ 2 = (1 + radical 3 / 2) C ^ 2
2Ab = radical 3 / 2
AB = root 3 / 2 * 1 / 2
Sina = radical 3 / 2
A=60°
B=30°

In the triangle ABC, it is known that a is equal to 30 degrees, AB is equal to 2, and AC is equal to root 3

AC:AB=√3:2=COS30°
right triangle
BC=AB/2=1
Angle B = 60 degrees
Angle c = 90 degrees

In △ ABC, a = 1, B= 3, ∠ a = 30 ° then ∠ B is equal to______

∵a=1，b=
3，∠A=30°
According to the sine theorem, a is obtained
sinA＝b
sinB∴sinB=
Three
2 ∠ B = 60 ° or 120 °
So the answer is: 60 ° or 120 °

If the middle angle of the high line is ∠ CAD = 20 ° a=

Because ad is a high line, and delta ADC is a right triangle
Therefore, ADC = 90 ° and CAD = 20 °
So ∠ C = 180 ° - 90 ° - 20 ° = 70 °
Since AE is an angular bisector, so ∠ BAC = 2 ∠ CAE = 2 ∠ BAE = 180 ° - ∠ B - ∠ C = 70 °
Then ∠ CAE = 35 ° is introduced
∠ead=∠cae-∠cad=35°-20°=15°
You can understand without drawing a picture

As shown in the figure, ad and AE are the angular bisector and high line of △ ABC, and ∠ B = 50 ° and ∠ C = 70 ° respectively, then ∠ EAD=______ .

∵∠B=50°，∠C=70°，
∴∠BAC=180°-∠B-∠C=180°-50°-70°=60°，
∵ ad is the angular bisector of ∵ ABC,
∴∠BAD=1
2∠BAC=1
2×60°=30°，
∵ AE is the high line of ∵ ABC,
∴∠BAE=90°-∠B=90°-50°=40°，
∴∠EAD=∠BAE-∠BAD=40°-30°=10°．
So the answer is: 10 degrees

In △ ABC, ∠ C = 110 °, B = 20 ° and AE is the bisector of ∠ BAC, then ∠ BAE=______ Degree

∵ in ᙽ ABC,  C = 110 °, B = 20 °, B + ∠ C + ∠ BAC = 180 °,
∴∠BAC=180°-∠B-∠C=50°．
AE is the bisector of ∠ BAC,
∴∠BAE=1
2∠BAC=25°．
Therefore, fill in: 25

As shown in the figure, in △ ABC: (1) Draw the high AD and center line AE on the side of BC (2) If ∠ B = 30 °, ACB = 130 °, calculate the degree of ∠ bad and ∠ CAD

(1) As shown in the figure:
（2）∵∠B=30°，∠ACB=130°，
∴∠BAC=180°-30°-130°=20°，
∵∠ACB=∠D+∠CAD，AD⊥BC，
∴∠CAD=130°-90°=40°，
∴∠BAD=20°+40°=60°．