Triangle inscribed in a circle | 3 Different Methods to Find the Radius of the Circle

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Күн бұрын

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@edwinpittomvils2570 2 ай бұрын

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

@jayprakashpatel2663 6 ай бұрын

3rd one is actually less time consuming and the good one

@sebrosacademy 2 ай бұрын

keep going . Great job

@prime423 6 ай бұрын

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!!

@MarieAnne. 8 ай бұрын

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)*

@giuseppemalaguti435 8 ай бұрын

R=ABC/4S=7*8*9/4*12√5=21/2√5

@Hot_Rock 8 ай бұрын

You need to prove your figures . Don’t just apply the laws or rules from math books.

@josleurs4345 8 ай бұрын

@@Hot_Rock depends just how far you will go ... in proof of known things :)

@soli9mana-soli4953 8 ай бұрын

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...

@jeanlismonde8718 8 ай бұрын

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

@devkrishna4769 8 ай бұрын

great video!

@RAG981 8 ай бұрын

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.

@mohabatkhanmalak1161 8 ай бұрын

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.

@holyshit922 8 ай бұрын

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

@lukaskamin755 6 ай бұрын

6:06 ша the point is on the same side from the horde AC. otherwise they add up to 180 degrees (or Pi radians)

@nunoalexandre6408 8 ай бұрын

Love it!!!!!!!!!!!!!

@prime423 7 ай бұрын

This is a very well problem with a formula for its solution. One needs to find the area and then apply formula.

@boryafod1370 4 ай бұрын

Через площадь S=sqrt[p(p-a)(p-b)(p-c)] = abc/(4R), где p - полупериметр.

@albertybermudez7454 8 ай бұрын

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

@adgf1x 8 ай бұрын

radius=5^1/2 unit

@tarunmnair 8 ай бұрын

R can be rationalised as 21.sqrt(5)/10 ...

@boriskurenkov4977 7 ай бұрын

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

@adgf1x Ай бұрын

R=abc/4.ar.triangle

@КатяРыбакова-ш2д 7 ай бұрын

2,1*V5

@josleurs4345 8 ай бұрын

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 ....

@НаиляНаипова 8 ай бұрын

А не проще применить теорему синусов

@murvetaykac7041 5 ай бұрын

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.