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PROPERTIES AND SOLUTIONS OF TRIANGLES

 Sine Rule: In any ∆ ABC,

where R is the radius of circumcircle of ∆ ABC
 

Cosine Rule: In any ∆ ABC,
a2 = b2 + c2 − 2bc cos A, b2 = c2 + a2 − 2ca cos B,
c2 = a2 + b2 − 2ab cos C 
                        or     

 Projection Rule: In any ∆ ABC,
    a = b cos C + c cos B, b = c cos A + a cos C, c = a cos B + b cos A
 Half Angle formulae: In ∆ABC, if a, b, c are the sides and 's' is the semi perimeter, then

 

 Napier's rule (or Tangent rule)

 Formulae corrected with area of a triangle

               
Where ∆ = Area of triangle ABC
             s = Semi perimeter
 m − n Theorem
If D is any point on the side BC of triangle ABC such that

and BD : DC = m : n
then (i) (m + n) cot = m cot α - n cot β
         (ii) (m + n) cot = n cot B - m cot C     

 Formulae corrected with Inradius, Exradii, Circumradius


 Orthocentre & Pedal triangle:
1. If D, E, F are the feet of perpendiculars drawn from the vertices A, B, C to the sides BC, CA and AB respectively, then ∆DEF is known as pedal triangle of ABC
2. The sides of pedal triangle are given by EF = a cos A, DF = b cos B, DE = c cos C
3. The angles of pedal triangle are 
    
4. The circumcircle of pedal ∆DEF is same as nine point circle of ∆ABC


        
5. Distance of orthocentre from vertices are given by  AH = 2R cos A, BH = 2R cos B, CH = 2R cos C
6. Distance of orthocentre from sides are given by  HD = 2R cos B cos C, HE = 2R cos A cos C, HF = 2R cos A cos B

Medians of a triangle and Centroid
     Let AD, BE, CF be the medians of
     ∆ABC concurrent at G (Centroid of ∆ABC)
 Appolonius Theorem


         
 The length of medians are given by (using Appolonius Theorem)


    

 G divides AD, BE and CF in the ratio 2 : 1

Distance of Circumcentre from vertices and sides
    SA = SB = SC = R
    SD = R cos A, SE = R cos B, SF = R cos C


 

Distance of Incentre from vertices & sides


 

Distance of Excentre from Vertices and Sides


    
Euler's line: If H, N, G, S represent orthocentre, nine point centre, centroid and circumcentre of ∆ABC then HN : NG : GS = 3 : 1 : 2. The line on which these points lie is known as Euler's line.
Distance between various centres associated with a triangle


     

CYCLIC QUADRILATERAL:
A quadrilateral is a cyclic quadrilateral if its vertices lie on a circle.

 
     
4. Ptolemy's Theorem: If ABCD is a cyclic quadrilateral, then
                                         AC.BD = AB. CD + BC. AD
i.e. in a cyclic quadrilateral, the sum of the products of the lengths of opposite sides is equal to the product of its diagonals.

 

REGULAR POLYGON:
A polygon is called a regular polygon if all its sides are equal and its angles are equal.


where 'a' is a length of side, 'n' is number of sides of polygon, 'R' is radius of circumscribing circle and 'r' is the radius of incircle of the polygon.
 

SOLUTION OF TRIANGLES:
     In a triangle there are six elements, viz., three sides and three angles. In plane geometry we have done that if three of the elements are given at least one of which must be a side, then the other three elements can be uniquely determined. The procedure of determining unknown elements from the known elements is called solving a triangle.

Solution of a right angled triangle
Case - I:
When two sides are given
Let the triangle be right angled at C,
then we can determine the remaining
elements as given in the following table.      

 

Case - II: When a side and an acute angle are given
     In this case we can determine the remaining elements as given in the following table.

 

SOLUTION OF TRIANGLES IN GENERAL
Case - I:
When three sides a, b and c are given
    In this case the remaining elements are determined by using the following formulae.

And A + B + C = 180°
Cosine formulae can also be used to find the angles.

 

Case - II: When two sides a, b and the included angle C are given


    In this case we use the following formulae:

Case - III: When one side a and two angles A and B are given
    In this case we use the following formulae to determine the remaining elements.
    A + B + C = 180o   C = 180− A− B


    
Case - IV: When two sides a, b and the angle A opposite to one side is given


   In this case we use the following formulae:

 

From (1), the following possibilities will arise:      
(a) When A is an acute angle and a < b sin A
In this case the relation sin B =   gives that sin B > 1, which is impossible.
Hence no triangle is possible.

(b) When A is an acute angle and a = b sin A
      In this case only one triangle is possible which is right angled at B.
(c) When A is an acute and a > b sin A
      In this case there are two values of B given by ,  say, B1 and B2 such that B1 + B2 = 180o.
Side c can be obtained by using c = 
Hence two triangles are possible. It is called ambiguous case.

Posted Date : 19-02-2021

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