The triangle whose largest angle is an acute angle must be an acute triangle () in a triangle, at least two angles are acute angles

Yes
A triangle whose largest angle is an acute angle must be an acute triangle (pair)
At least two corners of a triangle are acute (pair)

In the triangle ABC, if LGA - LGC = lgsinb = - LG Radix 2, and B is an acute angle, try to judge the shape of the triangle In the triangle ABC, if LGA - LGC = lgsinb = - LG Radix 2, and B is an acute angle, try to judge the shape of the triangle It seems to be useful for sine or cosine The answer is an isosceles right triangle

LGA - LGC = lgsinb = - LG radical 2
=>Sinc = 1 / 2
=>B = 45 degrees
Also sine theorem Sina / sinc = A / C = 1 / radical 2
=>Sinc = root 2 * Sina = sin (a + b) = (Sina + COSA) / Radix 2
=> sinA=cosA
=>A = 45 degrees
So C = 180-a-b = 90 degrees
So it's an isosceles right triangle

The value of tan15

=tan(60°-45°)
=(tan60°-tan45°)/(1+tan60°tan45°)
=(√3-1)/(√3+1)
=2-√3

If the difference between the area of a circle inscribed with a regular hexagon and an inscribed quadrilateral is 4, then the area of the circle is how much. It has to have a process. It is best to write it on paper and send it back, and give a good comment

In this paper, the radius 3R / 2 of the right triangle is the area of the inner circle 2R / 2

The area ratio of the inscribed square of the same circle to the circle, As the title

Let R be the radius of the circle,
Then the diagonal of the square is 2R
Square area (2R) 2 / 2 = 2R
Circle area π r
The area ratio is 2 R 2: π R 2 = 2: π

As shown in the figure below, the area of the square is 2 square decimeters. Find the area of the circle

π R2 = 3.14 × 2 = 6.28 (square decimeter);
A: the area of the circle is 6.28 square decimeters

If the circumcircle radius of a square and a regular hexagon are equal, what is the area ratio of the square to the regular hexagon? I want to explain

Let the radius of circumscribed circle be r
The square can be divided into four equal isosceles right triangles from the center of circumscribed circle and the vertex of square and regular hexagon. The length of the right angle side is the radius R; the regular hexagon is divided into six equal equilateral triangles, and the side length of the triangle is the radius R
so
Square area = 4 * (R ^ 2 / 2) = 2 * R ^ 2
The area of regular hexagon = 6 * (R * (√ 3) / 2 * r) / 2) = 3 * (√ 3) / 2 * R ^ 2
Area ratio of square to regular hexagon = (2 * R ^ 2): (3 * (√ 3) / 2 * R ^ 2) = 4:3 * (√ 3)

How to find the area ratio of inscribed square and circumscribed hexagon of a circle

The radius is r
Square area = 2R ^ 2
Hexagon area = 2sqrt (3) R ^ 2
Ratio = 1 / sqrt (3)

There are two squares. The side length of the big square is 4 meters longer than that of the small square, and the area of the large square is 80 square meters larger than that of the small square. There are two square faces

If the side length of the small square is x, then the side length of the large square is x + 4
Because area = side length * side length, so
X*X+80=(X+4)*(X+4)
By solving the equation, x = 8 can be calculated
The area of small square is 64 square meters, and that of large square is 144 square meters

It is known that the side length of a big square is 4 cm longer than that of a small square, and the area of a large square is 96 cm 2 larger than that of a small square

As shown in the figure:
Let the side length of the small square be a cm,
Then 4a+4a+4 × 4=96
8a=80
a=10
10 × 10 = 100 (square centimeter);
A: the area of a small square is 100 square centimeters

The triangle whose largest angle is an acute angle must be an acute triangle () in a triangle, at least two angles are acute angles