Using the sine and cosine tables, and if needed a calculator, do these problems.

Draw the picture shown in your notebook and explain how it was drawn.



Calculate the lengths of all three sides.

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.