The angle of a right angle AB C = 2 a C = 2

The angle of a right angle AB C = 2 a C = 2

Given the angle c = 90, ab = 2Ac
In a right triangle, the right angle to which 30 degrees are aligned is half the hypotenuse
So the angle B is 30 degrees
Angle a = 180 degrees - angle B - angle c
So the angle a = 60 degrees

In RT △ ABC, ∠ C = 90 ° AB = 2Ac, the degree of ∠ A and ∠ B is obtained

As shown in the figure, ∵ C = 90 °, ab = 2Ac,
∴∠B=30°，
∴∠A=90°-∠B=90°-30°=60°．

In the right triangle ABC, the angle c = 90 °, (1) given a, B, how to find the degree of angle a? (2) given a, C, how to find the degree of angle a (3) Given B, C, how to find the degree of angle a Formula or something, urgent

tanA=a/b
sinA=a/c
cosA=b/c

In the isosceles right triangle ABC, the angle c = 90 degrees, P is a point in the triangle ABC and PA: PC: Pb = 1:2:3

First, the triangle APC is rotated 90 degrees clockwise to make AC and CB coincide, and mark the past point P as P1, and the point from p 1 to triangle p 1 C is an isosceles triangle, because CP CP CP 1 is equal and the angle P C P1 is 90 degrees, the root of Zuoku theorem p 1 1 is 2 root sign 2, because the angle P P1 B of sufficiency theorem is 90 degrees, the angle APC is equal to the angle cp1b is 135 degrees

A right triangle ABC, where angle B = 90, the degree of angle a is three times of angle C. how many degrees are angles a and C?

A 67.5 B22,5

In the right triangle ABC, the angle c is a right angle, and the angle B: angle c = 1:4, find the degree of A

Angle a = 90-90 / 4 = 90-22.5 = 67.5

In the isosceles right triangle, the angle C in ABC is a right angle, and the distances between a point P and a, B and C are 3,1,2 respectively. Find the degree of angle BPC

PD vertical BC, PE vertical AC, vertical foot D, e
Then PD = CE, CD = PE
Let PD = CE = m, CD = pf = n, AC = BC = a
Then: PA ^ 2 = 3 ^ 2 = 9 = PE ^ 2 + AE ^ 2 = n ^ 2 + (A-M) ^ 2 (1)
PB^2=1=PD^2+BD^2=m^2+(a-n)^2 (2)
PC^2=2^2=4=PE^2+PD^2=m^2+n^2 (3)
(3)-(2) 3=2an-a^2
n=(a^2+3)/(2a) (4)
(3)-(1) -5=2am-a^2
m=(a^2-5)/(2a) (5)
a^2>5 (6)
(4) (5) generation (3)
a^4-10a^2+17=0
A ^ 2 = 5 + 2 * 2 ^ (1 / 2), or a ^ 2 = 5-2 * 2 ^ (1 / 2) (rounded off by (6))
So a ^ 2 = 5 + 2 * 2 ^ (1 / 2)
From cosine theorem: cos angle BPC = (Pb ^ 2 + PC ^ 2-bc ^ 2) / (2PB * PC)
=(1+2^2-5-2*2^(1/2))/(2*1*2)=-(2^(1/2))/2
So the angle BPC is 135 degrees

A right triangle ABC, ∠ C is a right angle. Given that the degree of ∠ A is one degree greater than that of ∠ B, how many degrees is ∠ a?

Because the triangle ABC is a right triangle and ∠ C is a right angle, so ∠ a + ∠ B = 90 degrees
It is also known that ∠ a - ∠ B = 1 degree
Therefore, a = 45
∠ B = 44.5 degrees

Given that the angle alpha is equal to 25 degrees and 42 minutes, the degrees of the remainder and complement of the alpha are calculated respectively

The remaining angle is 64 degrees and 18 minutes
The complementary angle is 154 degrees and 18 minutes

An angle is 25 ° larger than its remainder. Find the degree of the complementary angle and the complementary angle of this angle. Teach me before 23 o'clock

The remainder of this angle is: (90 ° - 25 °) / 2 = 32.5 °
The complementary angle of this angle is 180 ° - 25 ° - 32.5 ° = 122.5 °