In the triangle ABC, if the angle B = angle c = 15 ° AB = 2cm, CD is perpendicular to the extension line of AB intersecting AB at point D, then the length of CD is?

∵∠B=∠C=15°
∴AB=AC
∴∠DAC=∠B+∠ACB=30°
∵CD⊥AB
∴CD=1/2AC=1/2AB=1cm

In the triangle ABC, a ^ 2 + B ^ 2 = C ^ 2 + AB, find the size of angle C

By cosine formula
A ^ 2 = B ^ 2 + C ^ 2-2 * b * c * cosa ^ 2 is the square * and a is the angle between B and C,
that
The original form is
c^2=a^2+b^2-2ab*1/2
So COSC = 1 / 2, and because the angle c is an angle of the triangle, the angle c is at (0180),
So C = 60 degrees

In the triangle ABC, Tana = 1 / 4, tanb = 3 / 5. Find the size of angle c; if the side length of AB is √ 17, find the side length of BC

(1)
∵tanA=1/4,tanB=3/5
∴tan(A+B)
=(tanA+tanB)/(1-tanAtanB)
=(1/4+3/5)/(1-1/4*3/5)
=17/17
=1
∵C=180º-(A+B), 0º

Given the area of triangle ABC s = 1 / 4 (b square + C Square), find the size of three inner angles of triangle ABC Title as above. Online and so on! We haven't learned the mean inequality yet. This is only the problem of solving triangles

S=1/4(b^2+c^2)=1/2bcsinA
From the mean inequality, 1 / 2bcina = 1 / 4 (b ^ 2 + C ^ 2) > 1 / 2 bcsina = 1 / 4 (b ^ 2 + C ^ 2) > 1 / 2 bcsina = 1 / 4 (b ^ 2 + C ^ 2) > 1 / 2 bcsina = 1 / 4 (b=
1 / 4 (2BC) if and only if B = C
So Sina > = 1, so Sina = 1, so a = 90
Because when the equal sign holds, B = C, so B = C = 45
A=90 B=C=45

In △ ABC, ∠ A is two times of ∠ B, and the complementary angle of ∠ B is five times of the remainder of ∠ a. the degree of ∠ C is calculated

∵ A is 2 times of ∠ B, and  B's complementary angle is 5 times that of ∠ a,
∴∠A=2∠B，
180-∠B=5（90-∠A），
It is found that ∠ a = 60 ° and ∠ B = 30 °,
∴∠C=90°．

It is known that the complementary angle of △ ABC, ∠ A is three times of ∠ B, and the residual angle of ∠ B is 30 ° smaller than that of ∠ C. find out the size of the three internal angles of △ ABC?

The complementary angle of ∠ a = 180 degrees one (∠ B ＋∠ C) = 3 ∠ B
The remainder angle of ∠ B ＝ 90 degrees ～
By solving the above two equations, we can get ∠ B = 20 ∠ C = 100 ∠ a = 180-20-100 = 60

In 5 triangle ABC, if ∠ a = 2 ∠ B, and the complementary angle of ∠ B is 5 times of the residual angle of ∠ a, then the degree of ∠ a? Help me

According to the known conditions, the
180°-∠B=5（90°-∠A）…… （1）
∠A=2∠B…… （2）
Replace (2) with (1)
As a result, 180 ° - ∠ B = 5 (90 ° - 2 ∠ b)
180°-∠B=450°-10∠B
9∠B=270°
∠B=30°
Substitute ∠ B = 30 ° into (2)
It is concluded that: ∠ a = 60 °
Then ∠ a = 60 °

The complementary angle of a triangle is two times more than three degrees of its complementary angle. Find the complementary angle and complementary angle of this angle

Let the remainder angle be a
a+2a+3=180
a=59
So the remainder is 59
The complementary angle is 180-59 = 121

If the complementary angle of an angle is 30 ° smaller than that of its complementary angle, then this angle is______ .

If this angle is α, then its complementary angle is 180 ° - α, and the complementary angle of the remaining angle is 180 ° - (90 ° - α),
According to the meaning of the title, 180 ° - α = 180 ° - (90 ° - α) - 30 °,
A = 60 ° is obtained
So the answer is: 60 degrees

a. B and C are positive integers and satisfy the equation ABC + AB + AC + BC + A + B + C + 1 = 2004, then what is the minimum value of a + B + C?

ABC + AB + AC + BC + A + B + C + 1 = 2004ab (c + 1) + a (c + 1) + B (c + 1) + (c + 1) = 2004 (c + 1) (AB + A + B + 1) = 2004 (a + 1) (B + 1) (c + 1) = 2004 because a, B and C are all positive integers, then a + 1, B + 1, C + 1 are all positive integers, and they are greater than 1 and 2004 = 2 × 2 × 3 × 16

In the triangle ABC, if the angle B = angle c = 15 ° AB = 2cm, CD is perpendicular to the extension line of AB intersecting AB at point D, then the length of CD is?