draw a triangle ABC with side BC 7 cm Angle B 45 angle A 105 degree then construct a triangle whose sides are 4 by 3 times the corresponding side of angle abc
Answers
Analyzing the question:
We have a triangle ABC, with BC = 7cm, B = 45°, and A = 105°. Since BC is the only measurement we have, we'll have to construct two angles on B and C. We know that B = 45° but we don't know what angle has to be constructed on C. By using ASP of a triangle we get,
⇒ A + B + C = 180°
⇒ 105° + 45 + C = 180°
⇒ 150° + C = 180°
⇒ C = 180° - 150°
⇒ C = 30°
Now, we are asked to construct a triangle 4/3 times triangle ABC. Here 4/3 is greater than one, meaning it's bigger than the triangle ABC. So the triangle we'll construct will be outside triangle ABC.
Steps of Construction:
⇒ Draw a line BC of length 7cm.
⇒ Construct 45° on B and 30° and C.
⇒ Extend the line from B and C such that they meet at a point forming an angle of 105°, mark it as A.
⇒ Extend B to form a line BX such that it makes an acute angle with BC.
⇒ Make four equal divisions on BX, and name them B₁, B₂, B₃ and B₄.
⇒ Join B₃ to C.
⇒ Extend BC to Y.
⇒ Construct a line parallel to B₃C, from B₄, intersecting BY at C'.
Therefore, B₃C║B₄ by copying the angle ∠BB³C on B₄.
⇒ Extend AB.
⇒ Similarly construct a line parallel to AC from C', intersecting extended AB at A', Therefore CA║C'A'.
⇒ ∴ A'BC' is the required triangle.
