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.
find the unknown marked angles


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.
find the unknown angle 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°
  =  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.
find the unknown marked angle


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.
find the unknown marked angles

Opposite angles of a rhombus are equal.

In the Rhombus ABCD,
BCD = DAB = 51°

AD // BC and ADE is a straight line.
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.
find the unknown marked angle

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

Triangle AOD is an isosceles triangle.
OAD = ODA = 30°


ADB is a straight line.
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°