Question

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.

Answer 1

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Answer 2

∠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.

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.

Step 4

Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.

Step 5

Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.



Justification

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)

… (1)

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.

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