If alpha is equal to - 3, then which quadrant is the end of alpha?

If alpha is equal to - 3, then which quadrant is the end of alpha?

If α = - 3 is the radian unit, then the angle system is α = - 3 × 180 / π ≈ - 172 ° and it is obvious that the terminal edge of α is in the third quadrant

How to judge which quadrant alpha is

When angle a is between 0 and 90 degrees, it is the first quadrant; 90-180 degrees is the second quadrant; 180-270 degrees (or - 90-180 degrees) is the third quadrant; 270-360 degrees (or 0-90 degrees) is the fourth quadrant

Given that cos alpha is greater than 0 and Tan alpha is less than 0, find: (1) the set of angle alpha, (2) the quadrant where the terminal edge of alpha / 2 and alpha / 3 is located
(3) Try to judge the sign of Tan alpha / 2, sin alpha / 2, cos alpha / 2

Finding the quadrant of angle / N can divide the quadrants and divide each quadrant into N parts evenly. The range of angle / N starts from zero angle and is labeled 1, 2, 3 and 4 in turn. The quadrant number of angle / N is proposed, and the range of quadrant number proposed is the range of angle / n
COS is positive in quadrants one and four, sin is positive in quadrants one and two, and Tan is positive in quadrants one and three
Alpha is in the fourth quadrant, the set of quadrants is {2kx-x / 2,2kx}, K ∈ Z (use x instead of π)
The quadrants of alpha / 2 terminal edge are the second and fourth quadrants,
The quadrants of alpha / 3 are two, three and four,
(3) Tangent is negative, sine cosine may be positive or negative

If alpha is the third quadrant, which quadrant is half alpha?

Is that understandable? Understanding can be adopted

The sum of cosine and sine of an angle is greater than the sum of its cotangent and tangent. Which quadrant is the end of the angle

Classified discussion
first quadrant
All four trigonometric functions are positive
So there are
SiNx + cosx less than or equal to root 2
TaNx + Cotx greater than or equal to 2
So the first quadrant pass
Beta Quadrant
Tan and cot are both negative numbers
So the sum is negative,
As long as SiNx is greater than or equal to the absolute value of cosx
Similarly, the fourth quadrant
third quadrant
Both Tan and cot are positive
It is impossible that sin and COS are both negative
So, the third quadrant pass

Which quadrants are sine, cosine, tangent and cotangent positive and which quadrants are negative?

Sine: the first and second quadrants are positive, the third and fourth quadrants are negative cosine: the first and fourth quadrants are positive, the second and third quadrants are negative tangent: the first and third quadrants are positive, the second and fourth quadrants are negative. The formula is: "one all positive, two sine, three tangent, four cosine."

Given cosine a + sine a = 1 / 5, a is the tangent of the fourth quadrant

-3/4
(sinA+cosA)^2=1+2sinAcosA=1/25
(sinA-cosA)^2=1-2sinAcosA=49/25
cosA-sinA=7/5
cosA=4/5
sinA=-3/5

As shown in the figure: in the triangle ABC, D is a point on AB, ad = AC, and the middle AE on the side of BC intersects CD with F. prove: ab: AC = CF: DF

It is proved that if we make DM parallel BC and cross AE to m, then CF: DF = Ce: DM
If be = CE, then CF: DF = be: DM = AB: ad;
And ad = AC
So CF: DF = AB: AC

In the triangle ABC, the angle ACB is 90 degrees, the AE bisects the angle BAC, CD is perpendicular to AE, AB intersects D, AE intersects g, DF is parallel to BC, AC intersects F
Judge whether DC bisects angle FDE and explain the reason

Divide equally
Because AE bisectors BAC
So angle CAE = angle BAE
So in triangle ACG and triangle ADG
Angle BAE = angle CAE, Ag = AG, angle AGC = angle AGD,
So in triangle ACG is equal to triangle ADG
So AC = ad
So in triangle ace and triangle ade
AC = ad, angle CAE = angle BAE, AE = AE
So ace of triangle is equal to ade of triangle
So CA = de
So in triangle CDE, angle DCE = angle CDE
Because DF is parallel to CB
So angle CDF = angle DCE
So angle CDF = angle CDE
So DC bisector angle FDE

In the triangle ABC, ∠ ACB = 90 degrees, AC = BC, CD ⊥ AB, the perpendicular foot is D, e is the upper point of AC, f is the upper point of BC, and AE = BC, connecting DF and DF
If EF = 10, find the area of triangle EDF
In the triangle ABC, ∠ ACB = 90 degrees, AC = BC, CD ⊥ AB, the perpendicular foot is D, e is the point on AC, f is the point on BC, and AE = CF, connect de and DF, if EF = 10, calculate the area of triangle EDF
The answer I know is 25. What I want is the process

Because: triangle ABC is isosceles right triangle, so CD = ad = dB, so angle DCB = angle CAD = angle DBC = 45 degrees
So: Triangle CDF is equal to triangle ade: CF = AE, ad = CD, angle FCD = angle EAD
So: ed = FD, ADE = CDF
So: angle EDC + angle ade = angle EDC + angle CDF = angle EDF = 90 degree
So: Triangle EDF is isosceles right triangle
So: the area of triangle EDF is: EF * 1 / 2ef * 1 / 2 = 25