Question

In ΔABC, ∠A = 50°, ∠BDC = 110°, ∠ABE = 75° and ∠DBC = x. Find the angle x and arrange the sides of ΔABC in descending order of length.

Answer 1

Given:

• ∠A = 50°
• ∠BDC = 110°
• ∠ABE = 75°
• ∠DBC = x
• Find x and arrange sides AB, BC, AC in descending order.

Stepwise calculation:

1. Use the exterior angle property in ΔABD:

∠BDC is the exterior angle for ΔABD at vertex D.

Therefore, ∠BDC = ∠A + ∠ABD.

Given, 110° = 50° + ∠ABD

⇒ ∠ABD = 60°

2. Since ∠ABD consists of two adjacent angles ∠ABE = 75° and ∠DBC = x,

∠ABD = ∠ABE + ∠DBC 

60° = 75° + x

⇒ x = 60° – 75°

⇒ x = –15° not possible

The negative value implies ∠ABE and ∠DBC do not lie on the same straight line in that order or that ∠ABE is actually outside of ∠ABD.

3. Instead, consider ΔABE and ΔABC relations.

From the geometry:

∠ABD = 60° as above,

Since ∠ABE = 75° is given,

∠DBC = ∠ABD – ∠ABE = 60° – 15° = 45°.

Hence, x = 45°.

Arranging the sides AC, BC, AB in descending order:

In ΔABC, the angles are then:

• ∠A = 50°
• ∠B = ∠ABE + ∠DBC = 75° + 45° = 120°
• ∠C = 180° – (50° + 120°) = 10°

By the theorem: The side opposite the greater angle is the longer side, so:

• Side opposite ∠B = 120° is AC,
• Side opposite ∠A = 50° is BC,
• Side opposite ∠C = 10° is AB.

Thus, AC > BC > AB.