FANTASTIC JOB [1]
Just wanted to give u guys the formulas in properties of triangles.
First of course is a diagram, which indicates all the parameters of a triangle
I made it with paint...so pls forgive me if it is untidy
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7 Answers
DUE TO OBVIOUS REASONS, I WILL NOT BE ABLE TO GIVE U ALL THE THREE VARIANTS OF A FORMULA (IF THEY ARE PRESENT)
THE BEST WAY TO LEARN THESE FORMULAS IS TO NOICE THE PATTERN IN EACH OF 'EM AND REMEMBER THAT PATTERN
1) Sine Rule:
\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}
2) Cosine Rule:
cosA=\frac{b^2+c^2-a^2}{2bc}
3) Projection Formulae:
a=b cosC+c cosB
4) Napier's Analogy:
tan\left(\frac{B-C}{2} \right)=\frac{b-c}{b+c}cot\frac{A}{2}
5) Half Angle Formulae:
sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}
cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}
tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} (dividing sinA/2 by cosA/2)
6) Formulae for Area:
\Delta = \frac{1}{2}bc\; sinA=\frac{1}{2}ca\; sinB=\frac{1}{2}ab\; sinC
\Delta = \frac{abc}{4R}
\Delta = \sqrt{s(s-a)(s-b)(s-c)}
7) Formulae for R (circumradius):
R= \frac{abc}{4\Delta }
R= \frac{a}{2\; sinA}=\frac{b}{2\; sinB}=\frac{c}{2\; sinC }\; \; \; \left(from\; Sine\; Rule \right)
8) Formulae for r (inradius):
r= \frac{\Delta }{s}=(s-a)tan\frac{A}{2}=(s-b)tan\frac{B}{2}=(s-c)tan\frac{C}{2}
r= 4R\; sin\frac{A}{2}\; sin\frac{B}{2}\; sin\frac{C}{2}
9) Formulae For r1, r2 and r3( ex-radii):
r_1=s\; tan\frac{A}{2}
r_2=s\; tan\frac{B}{2}
r_3=s\; tan\frac{C}{2}
r_1= 4R\; sin(\frac{A}{2})\; cos(\frac{B}{2})\; cos(\frac{C}{2})
r_2= 4R\; cos(\frac{A}{2})\; sin(\frac{B}{2})\; cos(\frac{C}{2})
r_3= 4R\; cos(\frac{A}{2})\; cos(\frac{B}{2})\; sin(\frac{C}{2})
ALL THE BEST [4]
EASIEST OF ALL:
WITH USUAL NOTATIONS PROVE DAT \frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}
simple area of a triangle = (absinC)/2 = (bcsinA)/2 = (acsinB)/2
equating 1st and 2nd equalities .. a/sinA = c/sinC
equating 1st and 3rd equalities .. b/sinB = c/sinC
so a/sinA = b/sinB = c/sinC
PROOF for area of a triangle:
area of triangle = BC*AD/2 = AB*CF/2 = AC*BE/2
= (a*bsinC)/2 = (c*asinB)/2 = (b*csinA)/2
= (absinC)/2 = (acsinB)/2 = (bcsinA)/2
u can also prove it dis way.........
let
\vec{BC}=\vec{a}; \vec{CA}=\vec{b}; \vec{AB}=\vec{c}; by property \frac{1}{2}\left|\vec{a}X \vec{b}\right|;\frac{1}{2}\left|\vec{b}X \vec{c}\right|;\frac{1}{2}\left|\vec{c}X \vec{a}\right|; =\left|\vec{a}X \vec{b} \right|=\left|\vec{b}X \vec{c} \right| =\left|\vec{c}X \vec{a} \right|
therfore absin(pi-c)=bcsin(pi-a)=casin(pi-b)
= absinc=bcsina=casinb
divide by abc............................ n take reciprocals...