Answer :

Let us draw a triangle ABC with all the given angles.


BAC + ABC + ACB = 180°


45° + 65° + ACB = 180°


ACB = 180° - ( 45° + 65° ) = 70°



Since, We do not know the length of any side of triangle , So we cannot use the concept of perpendicular bisector to construct a circumcircle.


But we know that every triangle formed by a circumcentre with the two consecutive points is an isosceles triangle.


Let each internal angle of a triangle be divided into two angles by a point S’ as shown in fig.



Here,


α + β = 45° …(i)


α + γ = 65° …(ii)


β + γ = 70° …(iii)


Subtracting eq. (i) from (ii),


We get,


( α + γ) – (α + β) = 65° - 45°


γ - β = 20° …(iv)


Adding eq. (iii) and (iv),


We get,


(β + γ) + (γ – β) = 70° = 20°


2γ = 90°


γ = 45°


Substituting the value of γ in eq. (ii) and eq. (iii)


we get,


α + 45° = 65° and β + 45° = 70°


α = 20° and β = 25°


Steps of construction :


1.Mark the angle α on AX, such that RAX = α = 20°.


2. Cut the arc of length 2.5 cm on AR say BO. (r = 2.5 cm)


3. With O as a centre and radius equal to 2.5 cm cut the arc


on AX say OB.



4. With base as AB construct BAQ = 45° and ABR = 65°.


Let them meet at C.


5. Join BC and AC.



6. With O as a centre and radius = 2.5 cm draw a circle.



Thus we find that the circle with centre O is the circumcircle of Δ ABC.


We know that perpendicular from the circumcentre act as a perpendicular bisector.


Construct OF AB, OG AC and OH BC.



In right Δ OFA,


OAF = 20° and OA = r = 2.5 cm


AF = OA × cos 20°


AF = 2.5 × 0.939 cm (From table, cos 20° = 0.939)


AF = 2.347 cm = 2.35 cm


In Δ AOB


AF = BF


AB = 2(AF) = 2(2.35) = 4.7 cm


In right Δ OGC,


OCG = 25° and OC = r = 2.5 cm


GC = OC × cos 25°


GC = 2.5 × 0.906 cm (From table, cos 25° = 0.906)


GC = 2.265 cm


In Δ AOC


AG = GC


AC = 2(GC) = 2(2.265) = 4.53 cm


In right Δ OHB,


OBH = 45° and OB = r = 2.5 cm


BH = OB × cos 45°


BH = 2.5 × 0.707 cm (From table, cos 45° = 0.707)


BH = 1.767 cm


In Δ BOC


BH = HC


BC = 2(BH) = 2(1.767) = 3.53 cm


Thus AB = 4.7 cm, BC = 3.53 cm and CA = 4.53 cm


Length of all sides of triangle are 4.7 cm, 3.53 cm and 4.53 cm.


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