I used cosine rule to calculate the angle ACB = 71.8 degree. Let a = distance of C to the point P, the tangent of line BC to the circle. b = distance of B to the point Q, the tangent of line BC to the circle, c = distance of A to the point R, the tangent of line AC to the circle. Given that the line CO (center of the circle) bisects the angle of ACB, BO bisects angle CBA and AO bisects angle CAB, I got a = 3.5, b = 5.5 and c = 4.5. Angle OCP = 71.8 / 2 = 35.9 degrees. The radius r = a x tan 35.9 = 3.5 x 0.7237 = 2.53 units
Found the formula in the back of my copy Major-General Shortrede’s logarithmic Trig Function Tables (19th century). Didn’t know it’s called Heron’s formula. Shortrede uses it thus: SinA=(2/b*c)*area.
Hmm. so the answer turns out to be the triangle's area divided by S (the semi-perimeter of the triangle). Is it always that way? Because then we could solve for the radius in just a few seconds each time.
this is the first time i heard about that heron's formula. This algorithm is good for finding the maximum radius of a circle in a given triangle. I think. But square root in programming is quite slow.
Excelente explicación gracias por su tiempo y dedicación.
From Bogota D.C
Love It! Thank you for helping me solve my homework! Life just became much simpler!
Thank you, I'm happy to hear that!
I used cosine rule to calculate the angle ACB = 71.8 degree.
Let a = distance of C to the point P, the tangent of line BC to the circle.
b = distance of B to the point Q, the tangent of line BC to the circle,
c = distance of A to the point R, the tangent of line AC to the circle.
Given that the line CO (center of the circle) bisects the angle of ACB, BO bisects angle CBA and AO bisects angle CAB, I got a = 3.5, b = 5.5 and c = 4.5.
Angle OCP = 71.8 / 2 = 35.9 degrees. The radius r = a x tan 35.9 = 3.5 x 0.7237 = 2.53 units
Found the formula in the back of my copy Major-General Shortrede’s logarithmic Trig Function Tables (19th century). Didn’t know it’s called Heron’s formula. Shortrede uses it thus: SinA=(2/b*c)*area.
Nice one after long gap.
Many thanks Madam!
Hmm. so the answer turns out to be the triangle's area divided by S (the semi-perimeter of the triangle). Is it always that way? Because then we could solve for the radius in just a few seconds each time.
ya , r = area of triangle/semi perimeter of triangle , basically we call this inradius.
Heron's formula is one for the book for future reference, thank you!
so there is a equation to find radius of incircle, Area of triangle/half of the perimeter of triangle
the circle is called the incircle and the radius, inradius
This is very helpful. I've saved this off.
Exact form: r = (√231)/6
Like √8*9*10/ 6 ?
How do you get 231 and 6?
I got the same answer 😅
Is there any shorter method?
😂
I mean it's not hard and I can easily do it but it's too lengthy
After finding area of triangle using heron’s equation, you can use a simple equation: area of triangle equal to semi perimeter times radius.
@@drwzerThat’s what GPT4 does.
lol yes , r= area of triangle/semi perimeter of triangle !
this is the first time i heard about that heron's formula. This algorithm is good for finding the maximum radius of a circle in a given triangle. I think. But square root in programming is quite slow.
for anyone interested to implemented this in code, you can find r by:
r=2*total_triangle_area/|AC|+|AB|+|BC|
Very good!
Good lesson mam. Thanks a lot
It's my pleasure.
i love this video, it's very simple and helpfull
Thank you, I really appreciate your comment.
I used my excel formula, encoding only 3 sides measurement and l get the same result for radius.
Thanks for appreciating my method.
Ingenious...
Amazing solution, keep it up!
Merry Christmas
Merry Christmas to you too!
Tnx!❤
S=9*231^0.5/4 r=231^0.5/6
Heron e A = a b.c/4R
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