Solving Geometric Angles

Introduction for solving geometric angles:

The figure which consists of two rays with the same starting point and the angle which can be formed by the two arms on either side of the initial point and it is the vertex angle .There are different types of angles which based on their measuring degrees. Now we are going to see about the solving of geometric angles.


Types of solving geometric angles:


The different types of solving geometric angles is given by,

Right angle

Acute angle

Obtuse angle

Straight angle

Complementary angle

Supplementary angle

Right angle:

A right angle whose measure is 90°, is called a right angle.
Acute angle:

An acute angle whose measure is less than 90° is called an acute angle.

30°, 60°, 70° etc are all acute angles.

Obtuse angle:

An Obtuse angle whose measure is greater than 90° is called an Obtuse angle

120°, 135°, 140° etc are all Obtuse angle

Straight angle:

A Straight angle whose measure is 180° is called a Straight angle

Complementary angle:

A complementary angle is nothing but the sum of two angles measures 90° are called complementary angles.

30°, 60° are complementary angles .

Supplementary angle:

A supplementary angle is nothing but the sum of the two angles which measures 180° are called Supplementary angles.

120°, 60° are Supplementary angles. I have recently faced lot of problem while learning geometry tutoring online free, But thank to online resources of math which helped me to learn myself easily on net.


Example for solving geometric angles:


Ex1:

A geometric angle is 14° more than its complement. What is its measure?

Sol:

Let x° be the required angle.

Its complement=90°-x°

By the given condition:

90°-x°+14°=x°

2x°=104°

X°=52°

Hence required angle=52°

Ex2:

The measure of an geometric solving angle is double the calculate of its supplementary angle. Find its measure.

Sol:

Let the required angle =x°.

Its supplementary angle =180°-x°

By the given condition =2(180°-x°)

=360°-2x°

=120°

Hence required angle=120°

Ex3:

The two supplementary angles are the ratio2:3.Find the angles .

Sol:

Let the two angles in degrees be 2x and 3x

By the given condition=2x+3x=180°

5x=180°

X=36°

Hence the required angles are 2×36°=72° and

3×36°=108°

Geometry Practice Problems

Introduction for learning geometry problem answers:

The subdivision  of mathematics concerned with the properties of lines, curves and surfaces usually divided into pure, algebraic and differential geometry in accordance with mathematical techniques utilized.  The figures of two dimensions is called planes. learning Geometry problem answers is a module of math which involves about the study of shapes, lines, angles, dimensions, relative position of figures etc. it plays vital role in real time application like elevation, projection. Learning geometry problems answers provides many foundational skills and helps to build the thinking skills of logic, deductive reasoning, and analytical reasoning.Let us learn geometry problem answers. I like to share this What are Similar Triangles with you all through my article.


learning geometry problem answers:


A triangle has a perimeter of 56. If 2 of its sides are equal,then the  third side is 5 more than the equal sides, what is the length of the third side?

Solution:

Let y = length of the equal side
perimeter of triangle.

Perimeter of a  triangle = sum of all the 3 sides.
Plug in values of question.
56 = y + y + y + 5

Combine like terms
56 = 3y + 5

3y = 56 – 5
3y = 51
y =17

Note: the third side is 5 more than the equal sides.

So, the length of third side = 17 + 5 =22

Answer: The length of third side is 22.

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learning geometry problem answers:


The perimeter of a rectangle is 400 meters and its length is 3 times its width W. Find Width and Length, and the area of the rectangle.

Solution:

Use the perimeter formula to write.

2 L + 2 W = 400
"its length is 3 times its width W" into a mathematical equation as follows:

L = 3 W
We substitute L = 3 W in the equation 2 L + 2 W = 400.

2(3 W) + 2 W = 320
Expand and group like terms.

8 W = 400
Solve for W.

W = 50 meters
Use the equation L = 3 W to find L.

L = 3 W = 150 meters
Use the formula of the area.

Area = L x W = 150 * 50 = 7500 meters 2.

Geometry Questions

Introduction :

Geometry is a section in math, which deals with many aspects regarding shapes, figures. they involve with construction, study of their properties, area , volume, etc. They include study of solids too. Geometry deals with the entire concepts related to the shapes, solids, etc. Sample questions about the intersecting lines, area are in the following section.


Example geometry questions:


Here are few example geometry questions:

Geometry question 1:

Find the area of the triangle formed by (5,2), (-9,-3), (-3,-5)

Solution:

The formula for finding the area of the triangle formed by  (x1,y1), (x2,y2), (x3,y3)  is 1/2 | [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] |

Applying the formula,we get

1/2 | [5(-3 + 5) -9(-5 -2) -3(2 + 3)]|

1/2 |10 + 63 – 15|

1/2 |58|

Hence the area of the triangle is 29.


Geometry question 2:

Find the point which divides the line segment joining (-1,2) and (4,-5) in the ratio 3:2

The formula for a point which divides the line joining A(x1,y1) and (x2,y2) in the ratio l:m is

`((lx2 + my1)/( l + m) ,(lx2 + my1)/( l + m)) `

Applying the formula,

The point is

`((3 * 4 + 2 * (-1)) /( 3 + 2) , (3 * (-5) + 2 * 2)/(3 + 2))`

`((10)/(5),(-11)/(5))`

Hence the point is (2,-11/5)