Question

Срочно!!!!!!!!!!! Пліз допоможіть дам 20 балів Дано сторони трикутника 73, 52, 75. Знайти площу трикутника, радіуси описаного та вписаного кіл, висоту опущену на сторону довжини . 2. Дано сторони трикутника 27, 36, 45. Знайти кути трикутника в радіанах. 3. Дано сторони трикутника 3, 7, 8. Знайти всі бісектриси трикутника.

Answer 1

1) Даны стороны                 Периметр Р = 200

АВ (с) =   ВС (a) =   AC (b) =          Полупериметр р = 100

73         52            75    

Площадь S = √3240000 = 1800  

r = 18    

R = 39,54167    

Высоты    

h(a) = 2S/a = 69,23077    

h(b) = 2S/b = 48    

h(c) = 2S/c = 49,31507    

cos A = (b^2+c^2-a^2)/(2bc) = 8250 / 10950 = 0,75342  

A = arccos 0,75342 = 0,71754 радиан 41,1121 градуса

cos B = (a^2+c^2-b^2)/(2ac) = 2408 / 7592 = 0,31718  

B = arccos 0,31718 = 1,24805 радиан 71,5078 градуса

cos C = (b^2+c^2-a^2)/(2bc) = 3000 / 7800 = 0,38462  

C = arccos 0,38462 = 1,17601 радиан 67,3801 градуса

                            Сумма 180

2) Даны стороны

АВ (с) = ВС (a) = AC (b) =

27           36            45

cos A = (b^2+c^2-a^2)/(2bc) = 1458 / 2430 = 0,6  

A = arccos 0,6 = 0,9273 радиан 53,1301 градуса

cos B = (a^2+c^2-b^2)/(2ac) = 0  /1944 = 0  

B = arccos 0 = 1,5708 радиан 90 градуса

cos C = (b^2+c^2-a^2)/(2bc) = 2592 / 3240 = 0,8  

C = arccos 0,8 = 0,6435 радиан 36,8699 градуса

                    Сумма 180

3) Биссектрисы.

Используем формулу L(a) = √(bc(b+c+a)(b+c-a)/(a+b).

Подставим заданные длины сторон.      

L(a) = 24 18 4 √1728 =  41,56922/ 11= 3,77902

L(b) = 21 18 2 √756 =  27,49545 /10= 2,749545

L(c) = 56 18 12 √12096    = 109,9818/  15= 7,332121.

 

Answer 2

Triangle 1: Sides 73, 52, 75

To find the area of a triangle, we can use Heron's formula. Heron's formula states that the area of a triangle with sides of lengths a, b, and c is given by:

Area = sqrt(s * (s - a) * (s - b) * (s - c))

where s is the semi-perimeter of the triangle, calculated as:

s = (a + b + c) / 2

Let's calculate the area of the triangle with sides 73, 52, and 75:

Step 1: Calculate the semi-perimeter: s = (73 + 52 + 75) / 2 = 100

Step 2: Calculate the area using Heron's formula: Area = sqrt(100 * (100 - 73) * (100 - 52) * (100 - 75)) = 936

Therefore, the area of the triangle is 936 square units.

To find the radius of the circumcircle (circumscribed circle) of a triangle, we can use the formula:

Radius of circumcircle = (a * b * c) / (4 * Area)

where a, b, and c are the sides of the triangle, and Area is the area of the triangle.

Step 3: Calculate the radius of the circumcircle: Radius of circumcircle = (73 * 52 * 75) / (4 * 936) = 29.5

Therefore, the radius of the circumcircle of the triangle is 29.5 units.

To find the radius of the incircle (inscribed circle) of a triangle, we can use the formula:

Radius of incircle = Area / s

where Area is the area of the triangle, and s is the semi-perimeter of the triangle.

Step 4: Calculate the radius of the incircle: Radius of incircle = 936 / 100 = 9.36

Therefore, the radius of the incircle of the triangle is 9.36 units.

To find the height dropped onto the side of length 75, we can use the formula:

Height = (2 * Area) / 75

where Area is the area of the triangle.

Step 5: Calculate the height dropped onto the side of length 75: Height = (2 * 936) / 75 = 24.96

Therefore, the height dropped onto the side of length 75 is 24.96 units.

Triangle 2: Sides 27, 36, 45

To find the angles of a triangle in radians, we can use the Law of Cosines. The Law of Cosines states that for a triangle with sides of lengths a, b, and c, and angles opposite those sides A, B, and C respectively, the following equation holds:

c^2 = a^2 + b^2 - 2ab * cos(C)

Let's calculate the angles of the triangle with sides 27, 36, and 45:

Step 1: Calculate the angle opposite the side of length 45 (angle C): c^2 = a^2 + b^2 - 2ab * cos(C) 45^2 = 27^2 + 36^2 - 2 * 27 * 36 * cos(C) 2025 = 729 + 1296 - 1944 * cos(C) 2025 = 2025 - 1944 * cos(C) 0 = -1944 * cos(C) cos(C) = 0 C = π/2 radians

Step 2: Calculate the angle opposite the side of length 36 (angle B): b^2 = a^2 + c^2 - 2ac * cos(B) 36^2 = 27^2 + 45^2 - 2 * 27 * 45 * cos(B) 1296 = 729 + 2025 - 2430 * cos(B) 1296 = 2754 - 2430 * cos(B) 0 = -2430 * cos(B) cos(B) = 0 B = π/2 radians

Step 3: Calculate the angle opposite the side of length 27 (angle A): a^2 = b^2 + c^2 - 2bc * cos(A) 27^2 = 36^2 + 45^2 - 2 * 36 * 45 * cos(A) 729 = 1296 + 2025 - 3240 * cos(A) 729 = 3321 - 3240 * cos(A) 0 = -3240 * cos(A) cos(A) = 0 A = π/2 radians

Therefore, the angles of the triangle with sides 27, 36, and 45 are all π/2 radians.

Triangle 3: Sides 3, 7, 8

To find the bisectors of a triangle, we can use the Angle Bisector Theorem. The Angle Bisector Theorem states that in a triangle, the angle bisector of an angle divides the opposite side into segments that are proportional to the lengths of the other two sides.

Let's find the bisectors of the triangle with sides 3, 7, and 8:

Step 1: Calculate the length of the angle bisector opposite the side of length 3: Let the angle bisector opposite the side of length 3 divide the side into segments of lengths x and y. According to the Angle Bisector Theorem, we have the following proportion:

x / 7 = y / 8

Simplifying the proportion, we get:

8x = 7y

Since the angle bisector divides the side into two segments, we have:

x + y = 3

Solving the system of equations, we find:

x = 21/15 = 1.4 y = 12/15 = 0.8

Therefore, the length of the angle bisector opposite the side of length 3 is 1.4 units.

Step 2: Calculate the length of the angle bisector opposite the side of length 7: Let the angle bisector opposite the side of length 7 divide the side into segments of lengths x and y. According to the Angle Bisector Theorem, we have the following proportion:

x / 3 = y / 8

Simplifying the proportion, we get:

8x = 3y

Since the angle bisector divides the side into two segments, we have:

x + y = 7

Solving the system of equations, we find:

x = 21/25 = 0.84 y = 56/25 = 2.24

Therefore, the length of the angle bisector opposite the side of length 7 is 0.84 units.

Step 3: Calculate the length of the angle bisector opposite the side of length 8: Let the angle bisector opposite the side of length 8 divide the side into segments of lengths x and y. According to the Angle Bisector Theorem, we have the following proportion:

x / 3 = y / 7

Simplifying the proportion, we get:

7x = 3y

Since the angle bisector divides the side into two segments, we have:

x + y = 8

Solving the system of equations, we find:

x = 24/25 = 0.96 y = 56/25 = 2.24

Therefore, the length of the angle bisector opposite the side of length 8 is 0.96 units.

Therefore, the lengths of the angle bisectors of the triangle with sides 3, 7, and 8 are approximately 1.4 units, 0.84 units, and 0.96 units.