∠B = 45°, ∠A = 105°
Sum of all interior angles in a triangle is 180°.
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° − 150°
∠C = 30°
The required triangle can be drawn as follows.
Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.
Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.
Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.
The construction can be justified by proving that
In ΔABC and ΔA'BC',
∠ABC = ∠A'BC' (Common)
∠ACB = ∠A'C'B (Corresponding angles)
∴ ΔABC ∼ ΔA'BC' (AA similarity criterion)
In ΔBB3C and ΔBB4C',
∠B3BC = ∠B4BC' (Common)
∠BB3C = ∠BB4C' (Corresponding angles)
∴ ΔBB3C ∼ ΔBB4C' (AA similarity criterion)
On comparing equations (1) and (2), we obtain
This justifies the construction.