Long Answer TypeConstruct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the Δ ABC.
Steps of constructions:
1) Draw BC = 7 cm.
<2)
At B, construct ∠CBX = 45° and at C, construct
∠BCY = 180 - (45 + 105) = 30°
3) Let BX and CY intersect at A, &DElta; ABC so obtained is the given triangle.
4) Construct an acute angle Δ CBZ at B on opposite side of vertex A of δ ABC.
5) Mark-off four points ( greater of 4 and 3 in 3/4 points B1, B2, B3,B4 on BZ such that BB1 = B1B2 = B2B3 = B3B4.
6) Join B4 to C.
7) Draw B3C' parallel to B4C which meets BC at C'.
8) From C', draw C'A' parallel to CA meeting at A'.
Thus, A'BC' is the required triangle, each of whose side is times
the corresponding sides of
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