Construct a $△$ ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and $∠ B = 60^{\circ}$ .

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Solution

Steps of construction :

(i) Draw a line segment BC = 3.6 cm.

(ii) At B, draw a ray BX making an angle of $60^{\circ}$ and cut off BE = 4.8 cm.

(iii) Join EC.

(iv) Draw perpendicular bisector of CE which intersects BE at A.

(v) Join AC.

$△ A B C$ is the required triangle.

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Similar questions

Construct a $△ A B C$ in which $B C = 3.6 c m, A B + A C = 4.8 c m$ and $∠ B = 60^{\circ}$ .

Construct a $△$ ABC in which AC = 5 cm, and $∠ B A C = 60^{\circ}$ and AB – BC = 1.2 cm.

Δ A B C

B C = 3.6

(A C - A B) = 1.6

∠ A C B = 30^{\circ}

A B C

A B = 4.5 c m, A C = 5 c m

A

B C

3.6 c m

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