It is known that a, B, C are the three sides of the triangle ABC. If the square of a plus the square of B is equal to 10A + 8b minus 41, and C is The longest side of the triangle ABC, find the value range of C

It is known that a, B, C are the three sides of the triangle ABC. If the square of a plus the square of B is equal to 10A + 8b minus 41, and C is The longest side of the triangle ABC, find the value range of C

Because of the fact that (a? - 10A + 25) + (B? - 8b + 16) = 0 (a-5) 2 + (B-4) 2 = 0, so a-5 = 0 and B-4 = 0, a = 5, B = 4 according to the sum of the two sides of the triangle is greater than the third side, the difference between the two sides is less than the third side

In the triangle ABC, ABC is the opposite side of ABC respectively. It is known that the square of a minus (b minus C) is equal to BC. Find the angle A

From the cosine theorem, it is easy to know that the angle a is 60 or 120 degrees

In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively. It is known that a, B and C satisfy that the square of B equals a times C, and the square of a minus the square of B In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively. It is known that a, B and C satisfy that the square of B equals a times C, and the square of a minus the square of B equals a times C minus B times C. the size of a and the product of B times SINB divided by C are obtained

b^2=ac
A ^ 2-B ^ 2 = AC BC (a + B-C) (a-b) = 0 because a + b > C, A-B = 0, a = b
Substituting a = B into B ^ 2 = AC gives B = C, so the triangle ABC is an equilateral triangle
A = 60 degrees, bsinb / C = sin60 = radical 3 / 2

In the triangle ABC, a, B, C are the opposite sides of angle a, angle B and angle C respectively. The square of a = B times (B + C) to find positive, angle a is equal to 2 angle B

The cosine theorem is applied to ∠ A and ∠ B respectively. After substituting a ^ 2 = B ^ 2 + BC, the cosine ascending power formula can be used

In △ ABC, if a 2 + B 2 = 289, a 2-B 2 = 161, and C = 17, find the height on the longest side

From A2 + B2 = 289, A2-B2 = 161, a = 15, B = 14 are obtained,
∵ 152 + 142 = 172. According to the inverse theorem of Pythagorean theorem, the triangle is a right triangle,
The height of C is h, and the area formula s = 1
2ab=1
2ch，
∴h=15×14
17=210
17．

In the triangle ABC, if B = 60 degrees and the square of B is equal to AC, then what triangle ABC must be?

cosB=(a^2+c^2-b^2)/2ac
=(a^2+c^2-ac)/2ac
1/2=(a^2+c^2-ac)/2ac
ac=a^2+c^2-ac
a^2-2ac+c^2=0
(a-c)^2=0
A=c
That is ∠ a = ∠ C = (180-60) / 2 = 60
So ∠ a = ∠ C = ∠ B
ABC is an equilateral triangle

If s = the square of a - (B-C), then Tana / 2 is equal to

S = the square of a - the square of (B-C)
=a^2-b^2-c^2+2bc
=-(b^2+c^2-a^2)+2bc
=-2bccosA+2bc
=2bc(1-cosA)
And S = 1 / 2 * bcsina
Therefore, 2BC (1-cosa) = 1 / 2 * bcsina
(1-cosA)/sinA=1/4
tanA/2=1/4

Is the triangle ABC suitable for the condition that angle a equals half and angle B equals angle c?

Angle a equals half angle B equals angle C
Angle a + angle B + angle c = 180 degrees
So, angle a = angle c = 45 degrees, angle B = 90 degrees
Therefore, the triangle satisfying the condition is isosceles right triangle

In triangle ABC, angle a equals half angle B plus angle C

According to the meaning of the title ∠ a = 1 / 2 (∠ B + ∠ C) ①
The sum of the angles in the triangle is equal to 180 degrees
① Substituting ②, 1 / 2 (∠ B + ∠ C) + ∠ B + ∠ C = 180 ° is obtained
∠B+∠C=120°
∴∠A=60°

In triangle ABC, if angle a equals half angle B equals two thirds angle c, then what is angle c equal to

60 degrees
A=1/2B=2/3C
B=4/3C
A+B+C=2/3C+4/3C+C=3C=180
C=60