# Geometry: Finding Unknown Angles(Advanced Problems)

### Problem 1

In the figure below, ABC and DEF are triangles. AC = AB, AB // DF, BC // ED, AC // EF and ∠ CAB = 70°. Find ∠ x, ∠ y and ∠ z.

Δ ABC is an isosceles triangle.
 ∠ ACB = ∠ ABC = (180° − 70°) ÷ 2 = 55°

DF // AE and AD // EF
So, AEFD is a parallelogram.
 ∠ DAE = ∠ DFE = 70° ∠ x = 70°

DF // EB and DE // FB
So, DEBF is a parallelogram.
 ∠ EBF = ∠ EDF = 55° ∠ y = 55°

DE // CF and DC // EF
So, DEFC is a parallelogram.
 ∠ DCF = ∠ DEF = 55° ∠ z = 55°

### Problem 2

Sally makes a flag using two triangles, ABC and CDE, on a straight pole XY as shown below. Triangle ABC is an equilateral triangle. Triangle CDE is a right-angled isosceles triangle with EC = ED and CED = 90°. Find the unknown angle, p, between the two triangles.

Δ CDE is a right-angled isosceles triangle.
 ∠ ECD = ∠ EDC = (180° − 90°) ÷ 2 = 45°

Δ ABC is an equilateral triangle.
ACB = 60°

ECA is a straight line.
 ∠ ECD + ∠ p + ∠ ACB = 180° ∠ p = 180° − ∠ ECD − ∠ ACB = 180° − 45° − 60° = 75°

### Problem 3

The figure below shows a parallelogram AECF inside another parallelogram ABCD. AB = AC, AEC = 53° and FDC = 38°. Find t.

Opposite angles of a parallelogram are equal.
In the Parallelogram ABCD,
 ∠ ABC = ∠ ADC = ∠ FDC = 38°

Δ ABC is an isosceles triangle as AB = AC.

In the Isosceles Triangle ABC,
ABC = ACB = 38°

Sum of adjacent angles of a parallelogram is 180°.
In the Parallelogram AECF,
 ∠ EAF = 180° − ∠ AEC = 180° − 53° = 127°

Opposite angles of a parallelogram are equal.
 ∠ FCE = ∠ EAF = 127°

 ∠ t = ∠ FCE − ∠ ACE = ∠ FCE − ∠ ACB = 127° − 38° = 89°

### Problem 4

In the figure below, ABCD is a rhombus and ADE is a straight line. DAB = 51° and DCE = 42°. Find x and y.

Opposite angles of a rhombus are equal.

In the Rhombus ABCD,
BCD = DAB = 51°

Hence, DE // BC and BCED is a trapezium.

In the Trapezium BCED,
 ∠ x = 180° − ∠ BCE = 180° − ∠ BCD − ∠ DCE = 180° − 51° − 42° = 87°

BC = CD as all sides of a rhombus are equal.
Triangle BCD is an isosceles triangle.
 ∠ y = ∠ BDC = (180° − 51°) ÷ 2 = 64.5°

### Problem 5

ABC is a right-angled triangle with BAC = 90°. OA = OD = OF. EO // CA and DO // BC. Find x.

Triangle AOF is an equilateral triangle.
AOF = OFA = FAO = 60°
DAF = 90°
= 30°

Triangle AOD is an isosceles triangle.

ODB = 180° − ODA = 150°

DBEO is a trapezium.
DBE = 180° − ODB
DBE = 180° − 150° = 30°

ABC = DBE = 30°
In Triangle ABC,
 ∠ ACB = 180° − ∠ BAC − ∠ ABC = 180° − 90° − 30° = 60°

FCE = ACB = 60°
OECF is a trapezium.
 ∠ CEO = 180° − ∠ FCE = 120°

BEC is a straight line.
 ∠ x = 180° − ∠ CEO = 180° − 120° = 60°