PROPERTIES AND SOLUTIONS OF TRIANGLE
1. SINE FORMULAE :
asinA=bsinB=csinC=λ=abc2Δ=2R
where R is circumradius and Δ is area of triangle.
2. COSINE FORMULAE :
(b) cosB=c2+a2−b22ca
(c) cosC=a2+b2−c22ab
3. PROJECTION FORMULAE :
(b) ccosA+acosC=b
(c) acosB+bcosA=c
4. NAPIER'S ANALOGY (TANGENT RULE) :
(a) tan(B−C2)=b−cb+ccotA2
(b) tan(C−A2)=c−ac+acotB2
(c) tan(A−B2)=a−ba+bcotC2
5. HALF ANGLE FORMULAE :
s=a+b+c2= semi-perimeter of triangle.
(ii) sinB2=√(s−c)(s−a)ca
(iii) sinC2=√(s−a)(s−b)ab
(ii) cosB2=√s(s−b)ca
(iii) cosC2=√s(s−c)ab
(ii) tanB2=√(s−c)(s−a)s(s−b)=Δs(s−b)
(iii) tanC2=√(s−a)(s−b)s(s−c)=Δs(s−c)
=12bcsinA=12casinB=12absinC
=14√2(a2b2+b2c2+c2a2)−a4−b4−c4
6. RADIUS OF THE CIRCUMCIRCLE 'R' :
7. RADIUS OF THE INCIRCLE 'r' :
Point of intersection of internal angle bisectors is incentre and perpendicular distance of incentre from any side is called inradius 'r'.
8. RADII OF THE EX-CIRCLES :
Point of intersection of two external angle and one internal angle bisectors is excentre and perpendicular distance of excentre from any side is called exradius. If r1 is the radius of escribed circle opposite to angle A of △ABC and so on then :
(a) r1=Δs−a=stanA2=4RsinA2cosB2cosC2
(b) r2=Δs−b=stanB2=4RcosA2sinB2cosC2
(c) r3=Δs−c=stanC2=4RcosA2cosB2sinC2
9. LENGTH OF ANGLE BISECTOR, MEDIANS & ALTITUDE :
12√b2+c2+2bccosA=ma=12√2b2+2c2−a2
and βa=2bccosA2b+c,ha=acotB+cotC
Note that m2a+m2b+m2c=34(a2+b2+c2)
10. ORTHOCENTRE AND ORTHIC TRIANGLE
(a) Point of intersection of altitudes is orthocentre & the triangle KLM which is formed by joining the feet of the altitudes is called the orthic triangle.
(b) The distances of the orthocentre from the angular points of the ΔABC are 2RcosA,2RcosB, & 2RcosC
(c) The distance of orthocentre from sides are 2RcosBcosC, 2RcosCcosA and 2RcosAcosB
(d) The sides of the orthic triangle are a cosA(=Rsin2A), bcosB(=Rsin2B) and ccosC(=Rsin2C) and its angles are π−2A,π−2B and π−2C
(e) Circumradii of the triangles PBC,PCA,PAB and ABC are equal.
(f) Area of orthic triangle =2ΔcosAcosBcosC =12R2sin2 Asin2 Bsin2C
(g) Circumradii of orthic triangle =R/2
11. EX-CENTRAL TRIANGLE :
(a) The triangle formed by joining the three excentres I1,I2 and I3 of △ABC is called the excentral or excentric triangle.(b) Incentre I of △ABC is the orthocentre of the excentral ΔI1I2I3.
(c) △ABC is the orthic triangle of the ΔI1I2I3.
4RcosA2,4RcosB2 and 4RcosC2
and its angles are π2−A2,π2−B2 and π2−C2.
(e) II1=4RsinA2;II2=4RsinB2;II3=4RsinC2.
12. THE DISTANCES BETWEEN THE SPECIALPOINTS :
(a) The distance between circumcentre and orthocentre =R√1−8cosAcosBcosC
(b) The distance between circumcentre and incentre =√R2−2Rr
(c) The distance between incentre and orthocentre =√2r2−4R2cosAcosBcosC
(d) The distance between circumcentre & excentre is =R√1+8sinA2cosA2cosC2=√R2+2Rr1 & so on.
13. m-n THEOREM :
(m + n) cot θ = n cot B – m cot C
14. IMPORTANT POINTS:
(ii) If acosA=bcosB, then the triangle is isosceles or right angled.
(b) In Right Angle Triangle :
(ii) cos2A+cos2B+cos2C=1
(c) In equilateral triangle :
(ii) r1=r2=r3=3R2
(iii) r:R:r1=1:2:3
(iv) area =√3a24
(v) R=a√3
(ii) The orthocentre of right angled triangle is the vertex at the right angle.
(iii) The orthocentre, centroid & circumcentre are collinear & centroid divides the line segment joining orthocentre & circumcentre internally in the ratio 2:1, except in case of equilateral triangle. In equilateral triangle all these centres coincide.
15. REGULAR POLYGON :
Consider a 'n' sided regular polygon of side length 'a'.
(a) Radius of incircle of this polygon r=a2cotπn
(b) Radius of circumcircle of this polygon R=a2cosecπn
Perimeter =na=2nrtanπn=2nRsinπn
Area =12nR2sin2πn=nr2tanπn=14na2cotπn
16. CYCLIC QUADRILATERAL :
(a) Quadrilateral ABCD is cyclic if ∠A+∠C=π =∠B+∠C (opposite angle are supplementary angles)
(b) Area =√(s−a)(s−b)(s−c)(s−d), where 2s=a+b+c+d
(c) cosB=a2+b2−c2−d22(ab+cd) & similarly other angles.
(d) Ptolemy's theorem : If ABCD is cyclic quadrilateral, then AC⋅BD=AB⋅CD+BC⋅AD
17. SOLUTION OF TRIANGLE :
cosA=b2+c2−a22bc,cosB=c2+a2−b22ac & cosC=a2+b2−c22ab
Case-II : Two sides & included angle are given:
Let sides a,b & angle C are given then use tanA−B2=a−ba+bcotC2 and find value of A−B.....(i)
& A+B2=90∘−C2....(ii) c=asinCsinA.....(iii)
Case-III :
Two sides a,b & angle A opposite to one of them is given
(b) If a=bsinA & A is acute, then one triangle exist which is right angled.
(c) a>bsinA,a<b & A is acute, then two triangles exist
(d) a>bsinA,a>b & A is acute, then one triangle exist
(e) a>bsinA & A is obtuse, then there is one triangle if a>b & no triangle if a<b.
Note : Case-III can be analysed algebraically using Cosine rule as cosA=b2+c2−a22bc, which is quadratic in c.
18. ANGLES OF ELEVATION AND DEPRESSION
Let OP be a horizontal line in the vertical plane in which an object R is given and let OR be joined.
In Fig. (a), where the object R is above the horizontal line OP, the angle POR is called the angle of elevation of the object R as seen from the point O. In Fig. (b) where the object R is below the horizontal line OP, the angle POR is called the angle of depression of the object R as seen from the point O.
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