A triangle, its largest angle is 85 degrees, it must be a acute angle, b right angle, C angle

A triangle, its largest angle is 85 degrees, it must be a acute angle, b right angle, C angle

The biggest is the acute angle
So choose a

Radius formula of triangle inscribed circle

Let the three sides of the triangle be a, B, C, the area s, the circumscribed circle radius R and the inscribed circle radius R
Then s = 1 / 2 * (a + B + C) * r
R = 2S / (a + B + C)
Note: it is proved that if O is the center of the inscribed circle, then the triangle ABC is decomposed into three triangles: OAB, OBC and OAC, whose areas are 1 / 2 * Cr, 1 / 2 * AR and 1 / 2 * br. Then s = 1 / 2 * Ar + 1 / 2 * br + 1 / 2 * CR = 1 / 2 * (a + B + C) * r
S=abc/(4R)
R=abc/4S
Note: proof: obtained by sine theorem
a/sinA=2R
Sina = A / (2R)
S=1/2*bc*sinA
=1/2*bc*a/(2R)
S=abc/(4R)

How to prove the formula of radius of inscribed circle and trilateral relation of right triangle?

It is known that in RT △ ABC, ∠ C = 90 ° and inscribed circle ⊙ o cut AB, BC and Ca in D, e and f respectively
Verification: O radius = (a + B-C) / 2
It is proved that: ∵ o cuts AB, BC and Ca at points D, e and F,
According to the tangent length theorem, AE = AF, BD = BF, ﹤ AC + bc-ab = AE + CE + BD + cd-af-bf = CD + CE
∵ in quadrilateral CDOE,  C = ∠ CDO = ∠ CEO = 90 ° and OD = OE,
The quadrilateral CDOE is a square, CD = CE = OD,
⊙ o radius od = CD = (AC + bc-ab) / 2 = (a + B-C) / 2

Radius formula of inscribed circle of right triangle: r = (a + B-C) / 2 how is this formula derived?

Let RT △ ABC, ∠ C = 90 degrees, BC = a, AC = B, ab = C
The conclusion is that the radius of inscribed circle r = (a + B-C) / 2
There are generally two methods of proof:
Method 1
As shown in the figure, the center of the inscribed circle is O, and the three tangent points are D, e, F, connecting od and OE
Obviously, there are OD ⊥ AC, OE ⊥ BC, OD = OE
So the quadrilateral CDOE is a square
So CD = CE = R
So ad = B-R, be = A-R,
Because ad = AF, CE = CF
So AF = B-R, CF = A-R
Because AF + CF = AB = R
So B-R + A-R = R
Radius of inscribed circle r = (a + B-C) / 2
The diameter of inscribed circle L = a + b-c
Method 2
As shown in the figure, the center of the inscribed circle is O, and the three tangent points are D, e, F, connecting OD, OE, of, OA, ob, OC
Obviously, there are OD ⊥ AC, OE ⊥ BC, of ⊥ ab
So s △ ABC = s △ OAC + s △ OBC + s △ OAB
So AB / 2 = br / 2 + Ar / 2 + Cr / 2
So r = AB / (a + B + C)
=ab(a+b-c)/(a+b+c)(a+b-c)
=ab(a+b-c)/[(a+b)^2-c^2]
Because a ^ 2 + B ^ 2 = C ^ 2
So the radius of inscribed circle r = (a + B-C) / 2
The diameter of inscribed circle L = a + b-c

How to calculate? How to use formula? Integers 1,2,4,5,7,9 Question: 1. How many 3-digit numbers can be composed? 2. How many 3-digit numbers can be composed if these numbers are not repeated? 3. How many 3-digit numbers can be composed that are greater than 500 but not duplicated?

1. There are 6 optional digits in 100 digits, 6 optional digits in 10 digits, and 6 optional digits in 1 digit, so it is 6 * 6 * 6 = 216 3-digit
2. If you don't repeat numbers, look at the hundreds first, and you can select 6 numbers. After selecting 100 digits, only 5 digits are left in the ten digits. After selecting ten digits, there are 4 digits left in each digit. Therefore, it is 6 * 5 * 4 = 3 digits of 120 non repeating numbers
3. If the number is greater than 500 without repeating numbers, five, seven, and nine can be selected for the hundred digit. After the hundred digit is determined, there are five optional digits left in the ten digit number. After the ten digit number is determined, there are still four digits left for each of you to choose from. Therefore, it is 3 * 5 * 4 = 60 3 digits greater than 500 but without repeated digits

Some English math problems 1. The sum of number a and number B is 36.3. If the decimal point of number B is moved to the right by one place, the two numbers are equal. What is number a and number B? Rewrite sentences as required I have a good idea She's enjoying the music

1 I don't have a good idea. 2 is she enjoying the music no, she I

Define the function f (n) and calculate n + (n + 1) + (n + 2) + +(2n-1), the function return value type is double

f(n)=(3n-1)(n)/2=3n²/2-n/2

Define function Total (n) to calculate 1 + 2 +... + N, and define fact (n) to calculate

int total(int n){
return fact(n);
}
int fact(int n){
if(n==1)return 1;
else return n+fact(n-1);
}

The algorithm for finding the factorial of integer n (n > = 0) is as follows, and its time complexity is as follows: Int fact(int n) {if (n

B
This is a recursive path, we can see that every recursion n has a small scale, and the result is linear

How many zeros are there at the end of the factorial of 100? Accurate, can't let me calculate by myself! It should be 20 + 4 = 24!

Consider the number of factors 5, 20 multiples of 5 and 4 multiples of 25
20 + 4 = 24