Q. 44.5( 2 Votes )
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.
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.
Rate this question :






















Using the sine and cosine tables, and if needed a calculator, do these problems.
A triangle and its circumcircle are shown in the picture. Calculate the diameter of the circle.
Using the sine and cosine tables, and if needed a calculator, do these problems.
A triangle is made by drawing angles of 50° and 65° at the ends of a line 5 centimetres long. Calculate its area.
Kerala Board Mathematics Part-1Using the sine and cosine tables, and if needed a calculator, do these problems.
A circle is to be drawn, passing through the ends of a line, 5 centimetres long; and the angle on the circle on one side of the line should be 80°. What should be the radius of the circle?
Kerala Board Mathematics Part-1