In the triangle ABC, if a = 55, B = 16, and the area of triangle ABC is 220, 3, find ∠ C The ∠ C is calculated as 60 ° and 120 ° at last. How to exclude it~

In the triangle ABC, if a = 55, B = 16, and the area of triangle ABC is 220, 3, find ∠ C The ∠ C is calculated as 60 ° and 120 ° at last. How to exclude it~

Sorry, the picture is a little ugly
The area of the known triangle = 220 root sign 3
1 / 2 * 55 * H = 220 pieces 3
H = 8 √ 3 is obtained
SinC=h/AC= V 3/2 (the root of two is three)
So ∠ C = 60 ° or 120 °,
Because the triangle ADC is a right triangle
So it's 60 ° C

In triangle ABC, if angle a + angle B = 5 angle C In △ ABC, if ∠ a + ∠ B = 5 ∠ C, then ∠ 4=_____ Math exercise book, page 99, question 3

∠C=180/10=18

In the triangle ABC, angle a = 1 / 2, angle B = 1 / 5 angle C. what triangle is this triangle

Obtuse triangle a = 22.5 B = 45 C = 112.5

In the triangle ABC, 3 angles a > 5 angles B, 3 angles C ≤ 2 angles B

An obtuse triangle
Use the sum of the triangles 180 degrees
3A > 5B
2B >= 3C
Then 6A > 10B > = 15C
Add 15a and 15b to all three, and you get it
21A+15B > 15A+25B >= 15(A+B+C)=15*180
If B < 3a in 5, then
15*180

In the triangle ABC, we know that the angle a is greater than the angle B is greater than the angle c, and the angle a is equal to 2, the angle c is equal to 4, the edge a + the edge C = 8

By sine theorem
(a+c)/(sinA+sinC)=b/sinB
That is, 8 / (sin2c + sinc) = 4 / sin (180-3c)
Sin2c + sinc = 2sin3c
sin3C=sin(2C+C)=sin2CcosC+cos2CsinC=3sinC-4(sinC) ^3
2sinCcosC+sinC=6sinC-8(sinC)^3
2cosC=5-8(sinC)^2
2cosC=5-8[1-(cosC)^2]
8(cosC)^2-2cosC-3=0
COSC = - 1 / 2 or COSC = 3 / 4
cosC=(b^2+a^2-c^2)/2ab=[16+8(a-c)]/8a
a=24/5,c=16/5

In △ ABC, ∠ A is the smallest angle, ∠ B is the largest angle, and ∠ B = 4 ∠ a, then the value range of ∠ B is______ .

∵ A is the smallest angle,  B is the largest angle, and  B = 4 ∠ a,
∴∠A=1
4∠B，∠C=180°-∠A-∠B=180°-1
4∠B-∠B=180°-5
4∠B，
∵∠A≤∠C≤∠B，
∴1
4∠B≤180°-5
4∠B≤∠B，
∴3
2∠B≤180°，9
4∠B≥180°，
∴80°≤∠B≤120°．
Therefore, the answer is: 80 °≤∠ B ≤ 120 °

In △ ABC, ∠ A is the minimum angle, ∠ B is the largest angle, and 2 ∠ B = 5 ∠ A. if the maximum value of ∠ B is m ° and the minimum value is n °, then M + n=______ .

∵ 2 ∠ B = 5 ∠ a, i.e. ᙽ B = 5
2∠A，
∴∠C=180°-∠A-∠B=180°-7
2∠A，
And ? a ≤∠ C ≤∠ B,
∴∠A≤180°-7
2∠A，
The solution is ∠ a ≤ 40 °;
And ∵ 180 ° - 7
2∠A≤5
2∠A，
The solution is ∠ a ≥ 30 °,
∴30°≤∠A≤40°，
That is, 30 ° ≤ 2
5∠B≤40°，
∴75°≤∠B≤100°
∴m+n=175．
So the answer is: 175

In the triangle ABC, the angle a is larger than the angle B is larger than the angle c, and the angle a = 4, the angle c is used to find the range of the angle B

The maximum angle B is 180-90-90 divided by 4 = 180-115.5 = 64.5

In the triangle ABC, the angle a > the angle b > the angle c and the angle a = 4, the value range of the angle B can be obtained In the triangle ABC, angle a > angle b > angle c, and angle a = 4 angle C. find the value range of angle B

∠A=4∠C,∠B=180°-A-C
So ∠ C = 36-b / 5. ∠ a = 144-4b / 5
So 144-4b / 5 > b > 36-b / 5
That is 720-4b > 5B > 180-b
The answer is 30

In the triangle ABC, the angle B is equal to the angle c is equal to 15 degrees, ab = 2 cm, CD is perpendicular to the intersection of AB and the extension line of Ba is at point D, then what is the length of CD? We must have the correct answer. The more the better, the better,

Equal to 1
Because angle B equals angle c, AC = AB = 2 (equiangular to equilateral)
Because DAC + is not equal to the outside angle of the triangle
So CD = 1 / 2Ac
So CD = 1