An exterior angle (or external angle) is an angle formed by one side of a simple polygon and a line extended from an adjacent side.

In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles,

The sum of the internal angle and the external angle on the same vertex is 180°.

The sum of the measures of the external angles of a polygon ,one at each vertex is 360°.

Here angle 1,2 ,3 are external angles

1+2+3 = 360°.

Let’s see an example on this….from 10 grade math geometry

Find the value of x in the figure….

Since we have a straight line and a part of it is 80

Angle x=180-80

So x=100°

This also help solving slope of a line problems in algebra ii

Question:-

Out of forty students, 14 are taking English Composition and 29 are taking Chemistry. If five students are in both classes, how many students are in neither class? How many are in either class? What is the probability that a randomly-chosen student from this group is taking only the Chemistry class?

Answer:-

This online help with math gives the solution of the above problem.
Here the total number of students in class is what is Universal Set U ,which is given as 40 students

Here is the venn diagram template for the complete question

Two students are taking neither class.
There are 38 students in at least one of the classes.
There is a 24/40 = 0.6 = 60%  probability that a randomly-chosen student in this
group is taking Chemistry but not English.

If the initial term of an arithmetic progression is a₁ and the common difference of successive members is d, then the nth term of the sequence is given by:

and in general

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression.

Question:-

Find the 25th term of the following arithmetic progression
3, 6, 9, 12, 15, …

Answer:-

3, 6, 9, 12, 15, …

a = 3 , d = 3 , n = 25

Tn = a + (n − 1)d

T25 = 3 + (25 − 1)(3)

= 3 + 72

= 75

∴ the 25th term of the A.P. is 75.

For more help on this ,you can reply me.

Trigonometry is a branch of mathematics that deals with triangles, particularly those plane triangles in which one angle has 90 degrees (right triangles). Trigonometry deals with relationships between the sides and the angles of triangles and with the trigonometric functions, which describe those relationships.

This trigonometry help show us how to solve a problem using Tan formulas.

Question:-

 Solve 

  Cos2x=cosx

Answer:-

Lets write 

cos2x as 2 cos2x-1

So we have 

   2 cos2x-1=cosx

  Then subtract cosx on both sides

2cos2x-1-cosx=cosx-cosx

2cos2x-1-cosx= 0

Lets take cosx as y

So the equaton become

  2y2-y-1= 0

  (y-1)(2y+1)=0

  y-1=0 or 2y+1=0
                 -1
  So y = 1 or y= ------
                  2

Since y=cosx
                    -1
 Cosx=1   or  cosx= ----
                     2
We know that cos0=0 

So cosx=cos0

 x=0 ,

Similarly for the second equation 

 x=120 or 240

For more help on this ,Please reply me.

Topic : Length of Chords in a Circle of radius r

Using definite integrals we can calculate the length of chords with some constraints.

Question : In a circle of radius r, find the average length of the chords perpendicular to the diameter [-r, r].

Solution :

For more information and clarification related to the above topic you can find help at calculus help.

Topic : Height of a Cone

Formula for finding a Volume of cone = 1/3 * ￗ * r² * h

Question : Suppose sand is being poured onto a cone- shaped pile, beginning at time t = 0, at the rate of 29.4 cubic inches per minute. At t = 2 minutes the resulting cone has a diameter of 7″. As the sand continues to be poured, the cone is always similar to its original shape, but it grows in size. Find the height of the cone at t=2minutes(remember to include the base of the pile). Round your answers to the nearest hundredth

Solution :

Rate of increase of volume is 29.9 cubic inches

Hence in the first minute the volume becomes 29.4 cubic inches

So when t = 2 (second minute) Volume = 2 * 29.4 = 58.8 cubic inches

at t = 2 ; V = 1/3 * ￗ * r² * h ; r = 7/2 = 3.5″

So 58.8 = 1/3 * 3.14 * 3.5 * 3.5 * h

58.8 = 12.822 h

h = 58.8/12.822

h = 4.5858

h = 4.59″

Thus after two minutes the pile has a height of 4.59″

For more assistance please leave your comments and geometry help will get back to you.

Problem : The fourth term of a G.P. is 9 and the ninth term is 2,187. Find the first 4 terms of the G.P.

Solution :

Let the G.P. be a, ar, ar² , ar³, …

T₄ = ar³ = 9 ———- (1)

T₈ = ar⁸ = 2,187 —— (2)

(2) ¸(1) :   ar⁸ / ar³ = 2187/9

r⁵ = 243

r⁵ = 3⁵

r = 3

Substitute r = 3 into (1):  a(3³) = 9

a = 1/3

Therefore the first four terms of the G.P. are 1/3, 1/3(3), 1/3(3)², 1/3(3)³,

ie 1/3, 1, 3, 9

Problem : Solve (x+2)/(x-3) = 42/(x+7)

Solution :

Cross multiply,

(x + 2) (x + 7) = 42 (x – 3)

x + 2x + 7x + 14 = 42x -126

x + 9x + 14 = 42x – 126

   – 9x  – 14 = -9x – 14(adding -9x and -14 on both sides)

x = 33x – 140

x – 33x + 140 = 0

x – 28x - 5x + 140 = 0

x(x – 28) - 5(x + 28) = 0

(x-28)(x-5) = 0

x-28 = 0 or x-5=0

x = 28 or x = 5

Problem : Solve 3x -2y = 3 and x = y + 4

Solution :

3x -2y = 3 ——- (1)

x = y + 4 ——- (2)

Use equation (2) in (1) for substitution method

So, 3x -2y = 3

3(y+4) – 2y = 3

3y + 12 – 2y = 3

y + 12 = 3

Subtract 12 from both sides

y = – 9

Substitute y = -9 in equation (2)

x = y + 4

x = -9 + 4 

x = - 5

Question : Integrate e^1/x / x²

Solution :