Triangle inscribed in a circle | 3 Different Methods to Find the Radius of the Circle
ฝัง
- เผยแพร่เมื่อ 13 ก.ย. 2024
- Triangle inscribed in a circle | 3 Different Methods to Find the Radius of the Circle
MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : www.youtube.co...
Math Hunter : www.youtube.co...
--------------------------------------------------------------------------------
Become a member of this channel to get access of ULTIMATE MATH COURSE
Join the channel to become a member
/ @mathbooster
3rd one is actually less time consuming and the good one
Apply cosinus rule to ABC, this gives cos(B)=2/7 and sin(B)=3.sqrt(5)/7. Set B(0,0), O(4,h) then A(2,3.sqrt(5)). You get R out of R^2=|OB|^2=|OA|^2
keep going . Great job
R=ABC/4S=7*8*9/4*12√5=21/2√5
You need to prove your figures . Don’t just apply the laws or rules from math books.
@@Hot_Rock depends just how far you will go ... in proof of known things :)
A weapon in any Mathlete tool kit is the relationship between the radius of the inscribed or circumscribed circle to the Area of the triangle. Any well instructed Mathlete would use the applicable formula. It is an easy solve!!
Love it!!!!!!!!!!!!!
Через площадь S=sqrt[p(p-a)(p-b)(p-c)] = abc/(4R), где p - полупериметр.
I really liked the first method.
In second method, once we find cos θ = 11/21, we can then calculate
sin θ = √(1−(11/21)²) = √(1−121/441) = √(320/441) = 8√5/21
Then in △OBC, we can drop perpendicular from O to D on line BC.
Since △OBC is isosceles, OD bisect both BC and ∠BOC, and we get two congruent right triangles
△OBD ≅ △OCD with
OB = OC = R
BD = CD = 8/2 = 4
∠BOD = ∠COD = θ
In △OBD we get
sin θ = BD/OB
8√5/21 = 4/R
R = 4 * 21/(8√5)
*R = 21/(2√5)*
Thank you, enjoyed the three methods, especially the last one which is a formula with 'delta'. In practice like in engineering we come across these sort of problems often like the reverse case so this is helpfull.
Mas sencillo: buscar el coseno del ángulo en A y desde el centro perpendicular a 8 que divide en dos triángulos iguales y el ángulo que se opone a 4 también es igual al ángulo en A. Lo demás es sencillo
6:06 ша the point is on the same side from the horde AC. otherwise they add up to 180 degrees (or Pi radians)
This is a very well problem with a formula for its solution. One needs to find the area and then apply formula.
great video!
No need to use Cosine formula in last part ( from 13 min), nor find cos 2theta Halve the triangle OBC to give sin theta = 4/r. Sin theta is 8rt5/21, so find r.
tu calcules le demi périmètre p du triangle soit 12 et tu utilises la relation entre surface S et les longueurs des côtés a, b, et c
soit 4R.S= abc avec R le rayon du cercle circonscrit au triangle; tu calcules S avec la formule de Héron soit S = rac[p(p-a)(p-b)(p-c)]
tu trouves S = rac(5)/4 et tu sais que R = abc/4S soit R = rac(5)/4
Cosine law for cosine of an angle
Pythagorean trigonometric identinty to get sine from cosine
Sine law for radius
In second method cos(2theta) and second cosine rule is not necessary
We have isosceles triangle so if we know theta we know all angles in triangle BOC
If we drop perpendicular from O to BC we will bicect both angle BOC and side BC
just because we have isosceles triangle and central angle theorem
Now cos(90-theta) = 4/R
There are other possible methods I can't say if better than those shown, but
1) in your first method, once found that X = 2 than
BD = 2, DC = 6, AD = 3√ 5,
we can extend AD untill E, the intersection with the circle, and with intersecting chords theorem find DE in this way:
BD*DC = AD*DE
2*6 = 3√ 5*DE
DE = 4/5√ 5, and knowing that:
4r² = BD² + DC² + AD² + DE²
4r² = 2² + 6² + (3√ 5)² + (4/5√ 5)²= 441/5
r = 21/2√ 5
2) In your 3th method once known the area of ABC = 12√ 5 with Heron's formula, we can calculate sin(theta) in this way:
7*8*1/2*sin(theta) = 12√ 5
sin(theta) = 3/7√ 5
from sines law we know that:
2r = AC/sin(theta) = 9/(3/7√ 5)
r = 21/2√ 5 again...
second part of the second method would be more simple if one looks at the fact that center O lies on the perpendicular on the middle of relevant side of the triangle
R=abc/4.ar.triangle
2,1*V5
radius=5^1/2 unit
HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE
. Procedure
Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3
At the end we get the EXACT circumference of the inscribed circle
R can be rationalised as 21.sqrt(5)/10 ...
Why not cos rule , and then sin rule because Sine rule van be used tô find radius because a/sin = 2 R ... easy like that ....
А не проще применить теорему синусов
We can solve this problema using areas ofyringler formula.The first one is A=a.b.c/4R the second is A= (u.(u-a).(u-b).(u-c))1/2 2u=8+9+7=24 ..u=12 when we make all this operations we find R=21/2.5^1/2.Thank you telling us lots of formulas.good days.
Thales Theorem
Broooooo why enistein. Your boosting mathematics for god sakes