Question

construct a triangle ABC in which AB+BC+CA = 10cm. angle B= 60° and angle C= 45° .

Answer 1

Step 1: Draw a line segment PQ ( = AB + BC + CA ) = 10cm.
Step2: Draw an angle of 60° at P and an angle of 45° at Q.
Step 3: Bisect ∠P and ∠Q. Let the bisectors of ∠P and ∠Q intersect in A.
Step 4: Draw perpendicular bisector DE of AP which intersect PQ in B and FG of AQ which intersect PQ in C
Step.5: Join AB and AC. ∴ ΔABC is the required triangle


Hope it's help u

Answer 2

Answer:

A triangle ABC in which AB + BC + CA = 10 cm, ∠B = 60 °, and ∠C = 45 ° have been Constructed.

Step-by-step explanation:

Given Data :

AB + BC + CA = 10 cm, ∠B = 60 °, and ∠C = 45 °

To Find :

Construct a triangle using the given parameters.

To construct the triangle ABC, follow these steps:

• Draw a line segment XY equal to AB + BC + CA = 10 cm.

• Make the angle equal to ∠B = 60 ° from point X. Let the angle be ∠LXY.

• Make the angle equal to ∠C = 45 ° from point Y. Let the angle be ∠MXY.

• Bisect angles ∠LXY and ∠MXY. Let these bisectors intersect at point A.

• Make a perpendicular bisector of AX. Let it intersect XY at point B.

• Make a perpendicular bisector of AY. Let it intersect XY at point C.

• Join AB and AC.

• Triangle ABC is the required triangle.

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