Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
Given, ∠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°
we get ∠C = 30°
Construction Procedure:
The required triangle can be drawn as follows.
1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.
2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.
4. Join the points B3C.
5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.
6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.
7. Therefore, ΔA’BC’ is the required triangle.
Since the scale factor is 4/3, we need to prove
A’B = (4/3)AB
BC’ = (4/3)BC
A’C’= (4/3)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
A’B/AB = BC’/BC= A’C’/AC
A’B/AB = BC’/BC
= A’C’/AC = 4/3