Question

construct a triangle ABC in which AX, the altitude corresponding on BC is 3cm. Angle XAC =60 degree and Angle XAB = 45 degree measure BC.​

Answer 1

Answer:

1. Draw base BC of length 6.5 cm. Make an angle of 60

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60 ∘

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60 ∘ angle created and name it as A. Join A−C. △ABC is the required triangle.

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60 ∘ angle created and name it as A. Join A−C. △ABC is the required triangle.3. Draw a circle with centre as A and radius 3.5 cm. This is the locus of the points equidistant from A at a distance of 3.5 cm.

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60 ∘ angle created and name it as A. Join A−C. △ABC is the required triangle.3. Draw a circle with centre as A and radius 3.5 cm. This is the locus of the points equidistant from A at a distance of 3.5 cm.4. Draw the angle bisector of ∠ACB which will be the locus of points equidistant from AC and BC,

1. Draw base BC of length 6.5 cm. Make an angle of 60 ∘ from B.2. Mark an arc of radius 5 cm from B on the 60 ∘ angle created and name it as A. Join A−C. △ABC is the required triangle.3. Draw a circle with centre as A and radius 3.5 cm. This is the locus of the points equidistant from A at a distance of 3.5 cm.4. Draw the angle bisector of ∠ACB which will be the locus of points equidistant from AC and BC,5. Measure the length of XY which comes out to be 5 cm.

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