Is it better to serve hard to the inside or outside corner?

Personally, I feel that the inside corner is better, and it doesn't make much sense to give the outside corner if you serve strongly. Of course, this also depends on the situation, the situation of teammates and opponents, and the situation of the score have different strategies.

Either way, the serve should be matched with the tactics, the first zone sends the outside corner with a cut, pulls the opponent out of the court, and then shoots the forehand to death, the second zone with the outside corner with topspin, which is particularly useful against other people's backhands, and the second zone serves vigorously in the corner, and the net after the serve, which is very useful.

Personally, I think the inner corner is easier to send and the outer corner is too easy to go out of bounds.

The small man has inner horns, and the big man eats all.

I don't think there's a single trick in this world to eat everything that goes all over the world.

It's all different from person to person.

Look at your opponents, look at yourself...

Depending on the situation, you must observe the opponent's position before serving, and decide whether to serve the inside corner, the middle or the outside corner, the flat shot or the side spin, or the topspin according to the opponent's position!

Normally, the following three situations are the following three situations, the flat shot is sent to the inside corner, the topspin ball or flat shot is sent to the close position, and the side spin is sent to the outside corner, which is a common situation! But it also depends on the position of the other party!

For example, if your opponent is standing close to the inside corner, you can use a flat shot or a sidespin ball to the outside corner, or you can use a topspin ball to get close! This can force the opponent to dodge to catch the ball, or make a mistake.

The sum of the inner angles of the triangle is 180°

So the degree of this outer angle is 45°

So the degree of the inner angle adjacent to this outer angle is 180°-45°=135°

(n-2)*180-a+(180-a)=600(n-1)*180=600+2a

n-1=(600+2a) 180=3+(60+2a) 180 because 0 a 180, then 0 2a 36060 60+2a 420

Then 1 3 (60+2a) 180 7 3=2 1 3 takes the integer (60+a) 180=1,2

n-1=3+1=4

n=5 ，a=60

or n-1=3+2=5

n=6，a=150

1500-180=1320 (subtract 1500 from the degree of the outer angle and the inner angle adjacent to it).

Let the side length of this polygon be n, and the sum of the inside angles of this polygon minus 1320 should be less than 180 degrees.

0<（n-2)*180-1320<180

0<180n-360-1320<180

1680<180n<180+360+132028/3n<31/3

n is an integer, i.e. n=10

It is a 10-sided shape.

One inner angle + outer angle = 180 and the rest of the inner angles and + one inner angle + outer angle = 540 + 60 = 600

The rest of the inner angles and +an inner angle must be an integer multiple of 180)

The rest of the internal angles and +one internal angle = 540 ( n-2)180 = 540 n = 5

There seems to be a problem with this problem, using the conventional method to solve that the outer angle is 60°, which is a 5-sided shape, but then the sum of the outer angles becomes 300°, but no matter how you check the process, there will be no mistakes, what hell!!

Let the number of sides of this polygon be n

600<180*(n-2)<600+180600<180n-360<780

960<180n<1140

n<,n=6,

Solution: Let the number of sides of this polygon be n, because the degree of an outer angle is between 0 180, so 600-180<(n-2)*180<60013 3 Because n is an integer, n=5

That is, the number of sides of this polygon is 5, so the sum of its inner angles is 540 degrees, and the outer angles are 60 degrees.

The sum of the outer angles of one of the inner corners of the polygon and the other interior angles is 600°

The sum of the outer angles of one of the inner corners of the polygon + the sum of the other internal angles = 540 + 60 The outer angles are less than 180 and the outer angles = 60

360 60=6 polygons.

Let the number of sides of this polygon be n, then its inner angles and =(n-3)*180° Let the vertices of this angle be a, and its two adjacent vertices are b and c, slightly extending ca to d, that is, bac is the inner angle, and bad is the outer angle.

Connect b and c to divide the original polygon into a triangle and a polygon composed of n-1 edges, so: (n-1-3)*180° + bad+ abc + acb = 600°

Whereas bad= abc+ acb

So: (n-4)*180°+2* bad=600° because: 0 2* bad 360° so:

600°-360° (n-4)*180° 600°So: n-4

So: n=6,7

When n=6, the outer angle bad=120°

When n=7, the outer angle bad=30°

Let the inner angle of the polygon be x (degree omitted), then there is x-(180-x)=90, and 2x=270Therefore x = 135 degrees. Let this polygon be an n-side, then 180(n-2)=135n, and n=8 can be solved

Therefore, it is a positive edge, and the sum of the inner angles is 1080 degrees.

The sum of the outer angles is 360, which can be obtained from the title, and the inner angle is 360*2=720

180*（n-2）=720

n=6 so it is hexagonal.

Let the inner angle be x, x+1 4x=180, x=144, 1 4=36,360 divided by [the division sign will not be played] 36=10, 144x10=1440, answer: the number of sides is 10, and the sum of the inner angles is 1440 degrees.

A closed figure consisting of three or more line segments on the same plane and not on the same line that are connected one after the other and do not intersect is called a polygon. A figure composed of multiple line segments on different planes that are connected sequentially and do not intersect is also called a polygon, which is a polygon in a generalized sense.

In mathematics, the sum of the inner angles of a triangle is 180°, and the sum of the inner angles of a quadrilateral (polygon) is 360°. And so on, add an edge, and the sum of the internal angles is 180°.

The sum of the internal angles is: (n 2) 180° The number of internal angles of a regular polygon is: (n 2) 180° n For example, the sum of the internal angles of a triangle is the sum of the three internal angles, and an internal angle is any one of the corners.

Solution: Knowing from the meaning of the question.

In a polygon, each inner angle is equal, and each outer angle is equal to 1/4 of its adjacent inner angles

Let the variable of the polygon be n

The sum of the inner angles of a polygon is 180o (n 2), and the outer angle of any polygon is 360o ?×180o×（n－2）／n＝360o／n∴n＝10

The number of sides of this polygon is 10

The sum of the inner angles is 180o (n 2) 180o (10 2) 1440o I wish you success in your studies!

Let the inner angle be x

x+1/4x=180x=144

360 divided by [the division sign will not be played] 36 = 10

144x10=1440

Answer: The number of sides is 10, and the sum of the inner angles is 1440 degrees.

There is not a process, and a 5-sided shape, 540 degrees seem to be.