Given that the sum of the two right angles of a right triangle is equal to 8, what is the maximum area of the right triangle?

Given that the sum of the two right angles of a right triangle is equal to 8, what is the maximum area of the right triangle?

Let the right side of a right triangle be x, then the other right side is 8-x. the area of a right triangle is s
According to the meaning of the title, we get
S=1
2x（8-x）（0＜x＜8），
Formula, get
S=-1
2（x-4）2+8；
When x = 4, that is, when the two right angles are 4, the area of the triangle is the largest, and the maximum area is 8

Given that the sum of two right triangle is equal to 8, how many are the two right sides, the area of this right triangle is the largest, what is the maximum value?

Let one right angle side be x and the other right angle side be 8-x
S=1/2x(8-x)
=-1/2x*2+4x
Finding the maximum value of quadratic function
Maximum (4ac-b) / 4A = 8
Or: S = 1 / 2x (8-x)
=-1/2x*2+4x
=-1/2（x*2-8x）
=-1/2（x*2-8x+16-16）
=-1/2（x-4）*2+8
When x = 4, the maximum value is 8

Given that the sum of two right angles of a right triangle is equal to 8, what is the maximum area of the right triangle Junior three mathematics Requirement calculation process

If a right angle side is x, then the other angle is 8-x,
S=1/2X(8-X)
=-1/2(X^2-8X)
=-1/2(X^2-8X+16-16)
=-1/2(X-4)^2+8,
When x-4 = 0, that is, x = 4,
Smax = 8

Given that the sum of the two right angles of a right triangle is equal to 8, what is the maximum area of the right triangle?

Let the right side of a right triangle be x, then the other right side is 8-x. the area of a right triangle is s
According to the meaning of the title, we get
S=1
2x（8-x）（0＜x＜8），
Formula, get
S=-1
2（x-4）2+8；
When x = 4, that is, when the two right angles are 4, the area of the triangle is the largest, and the maximum area is 8

Let ABC be the opposite side of angle a, angle B and angle C in the triangle, and the area of the triangle is the square of a + the square of (B-C), and then find the tangent value of a / 2

s=a²+（b-c)²
tanA=2(tana/2)/(1-tan²a/2）
s=bcsinA/2,
sinA=2[a²+（b-c)²]/bc
a²=b²+c²-2bccosA
cosA=（b²+c²-a²）/2bc
tanA=2[a²+（b-c)²]/bc/（b²+c²-a²）/2bc
=4(a²+b²+c²-2bc)/（b²+c²-a²）
The title seems to be really wrong

In triangle ABC, AC = 100, tangent of angle a = 1, tangent of angle c = 2, calculate area of BC and triangle ABC Baidu knows not to, I can't understand

If the angle a = 45 degrees, then BD = ad = 2CD, then BC can be obtained according to Pythagorean theorem, and then the area is 0.5 * BD * AC

Given an angle and opposite side of a triangle, the problem of finding the maximum area of a triangle Let a be an inner angle of the acute triangle ABC, BC = 2, cos 2A = - 7 / 25, and find the maximum area s of triangle ABC How to do this? I tried to use the sine theorem to get the relationship between S and angle B. I can get the result, but that is too complicated. Do you have any simple method? That's what I did. I mean, is there no easier way?

Here is a geometric method for your reference
Theorem: if the BC side length of the triangle ABC is constant and ∠ A is equal to the known angle (i.e., the size is constant), then the trajectory of point a is an arc with BC as the chord and the circumference angle of the circle is equal to the known angle. (in fact, it is about two circular arcs symmetrical about BC. For this problem, due to the symmetry, we can only care about one of the arcs)
Because the problem △ ABC is an acute angle triangle, cos2a = - 7 / 25 can prove that the size of ∠ A is determined, so point a is on an arc with BC as the chord. Obviously, when a moves along the arc to the vertical plane of BC, the height on BC gets the maximum value, so the area of triangle ABC is also the maximum value, So the maximum value of △ ABC is (1 / 2) · 2.2 = 2

In △ ABC, ab = 1, BC = 2, then the value range of tangent of C is Process~~~~~~~~~~

In △ ACB, a semicircle is drawn with point B as the center and ab as the radius, and the position of AB edge changes to change ∠ C. when AC cuts semicircle B, ab (radius) ⊥ AC,
The maximum value of angle c forms a right triangle BAC, ∠ BAC = 90 ° and BC = 2 hypotenuse and ab = 1 right angle side. In other words, Tan 0 ° < Tan C ≤ Tan 30 ° is Tan 0 ° < Tan C ≤ Tan 30 ° and the value range of tangent of C is 0 < Tanc ≤√ 3 / 3, (√ radical)
I don't know if you are satisfied or not, hope to give you some help

Observe the value range of tangent curve TaNx = 0 X

K π and process?

In △ ABC, ∠ ACB = 90 ° with a circumference of (5 + 2 roots, 3) cm, and the center line CD = 2cm on the oblique edge In △ ABC, if ∠ ACB = 90 ° and the perimeter is (5 + 2, 3) cm, the median line CD on the oblique side is 2 cm, then the area of RT △ ABC is

∵ the center line on the hypotenuse = half of the hypotenuse
Ψ bevel C = 4cm
a+b=5+2√3-4=1+2√3
Square on both sides
a^2+b^2+2ab=13+4√3
According to Pythagorean theorem
a^2+b^2=c^2=16
16+2ab=13+4√3
2ab=4√3-3
The area of RT △ ABC = AB / 2 = √ 3-3 / 4