Construct a triangle $A B C$ in which $B C = 7 c m$ , $∠ B = 75^{o}$ and $A B + A D = 13 c m$ .

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Solution

1. Draw $B C = 7 c m$
2. Draw angle $75^{\circ}$ at $B$ and cut an arc of $B D = 13 c m$ .
3. Join $C D$ .
4. Draw a perpendicular bisector of $C D$ which meets $B D$ at $A$ .
5. Join $A C$ .
$A B C$ is the required triangle.

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△ A B C

B C = 7 c m ∠ B = 75^{o}

A B + A C = 13 c m

Construct a triangle ABC in which $B C = 7 c m, ∠ B = 75^{\circ} a n d A B + A C = 13 c m$

△ A B C

B C = 7 c m, ∠ B = 75^{o}

A B + A C = 12 c m

Construct a triangle $A B C$ in which $B C$ equal to $7 c m$ , Angle $B$ equal to $75$ degree and $A B + A C$ equal to $13 c m$ .

P Q R

Q R = 7 c m, ∠ Q = 75^{o}

P Q + P R = 13 c m

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