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 $13 c m$ .

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Solution

Step 1 : Draw a straight line $B C$ with length $7 c m$ . Mark the left end as $B$ and right end as $C$ .
step 2 : construct a ray $B M$ from point $B$ with an angle $75^{\circ}$ with respect to line $B C$ .
Step 3: Mark an arc of length $A B + A C = 13 c m$ on $B M$ from centre $B$ that arc intersect $B M$ at the point $D$ .
Step 4: Join $C D$
Step 5: Now, draw a perpendicular bisector of $C D$ that intersect $B D$ at the point $A$ .
Step 6: Join $A C$ .

Therefore, $A B C$ required triangle.

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Construct a triangle ABC in which BC equal to 7 cm, Angle B equal to 75 degree and AB+AC equal to 13 cm.

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 $13 c m$ .