9 Find the value
of x,y,z in
the
following figure
X Х
60°
Z
Answers
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step by step explanation:
Solution (i):
ABCD is a rectangle. We know that the diagonals of the rectangle are equal and bisect each other.
⟹BD=AC and AO=BO=OC=OD
In △ABO,
∠OAB=∠OBA=50
∘
---Angles opposite to the equal sides in a triangle are equal.
In △AOB,
∠AOB+∠OAB+∠OBA=180
∘
---sum of interior angles of a triangle
∴∠AOB=x=180
∘
−50
∘
−50
∘
=80
∘
Solution (ii):
ABCD is a rectangle. We know that the diagonals of the rectangle are equal and bisect each other.
⟹BD=AC and AO=BO=OC=OD
∠OCD+∠DCE=180
∘
---Angles on a straight line
∠ODC=180
∘
−146
∘
=34
∘
In △DOC,
∠ODC=∠OCD=34
∘
---Angles opposite to the equal sides in a triangle are equal.
In △DOC,
∠CDO+∠DCO+∠DOC=180
∘
---sum of interior angles of a triangle
∴∠DOC=180
∘
−34
∘
−34
∘
=112
∘
∠DOC=∠AOB=x=112
∘
---Vertically Opposite angles
∠DCO=∠OAB=y=34
∘
---Alternate angles
Solution (iii):
ABCD is a Square
∠ABC=∠BCD=∠CDA=∠DAB=90
∘
---interior angles of a square are equal to 90
∘
BCE is an equilateral triangle
∠EBC=∠BCE=∠BEC=60
∘
∠ECD=∠BCD+∠BCE=90
∘
+60
∘
=150
∘
In △CDE
CE=CD ⟹∠CED=∠CDE ---Angles opposite to the equal sides in a triangle are equal.
∠CDE+∠CED+∠DCE=180
∘
---sum of interior angles of a triangle
2∠CED=180
∘
−150
∘
=30
∘
∠CED=15
∘
Therefore, ∠BEC=∠BED+∠CED
60
∘
=x+15
∘
x=60
∘
−15
∘
=45
∘
Solution (iii):
BCE is an equilateral triangle
∠EBC=∠BCE=∠BEC=60
∘
ABCD is a Square
∠ABC=∠BCD=∠CDA=∠DAB=90
∘
---interior angles of a square are equal to 90
∘
∠BCD=∠ECD+∠ECB=90
∘
∠ECD=90
∘
−60
∘
=30
∘
In △DEC,
CE=CD ⟹∠CED=∠CDE=y ---Angles opposite to the equal sides in a triangle are equal.
∠CDE+∠DCE+∠DEC=180
∘
---sum of interior angles of a triangle
∴y+y=180
∘
−30
∘
=150
∘
y=75
∘
∠ADC=∠ADE+∠EDC=90
∘
75
∘
+x=90
∘
x=15
∘
y+z+∠CEB=360
∘
---Central Angle measures 360
∘
z=360
∘
−75
∘
−60
∘
=225
∘
Solution (iv):
∠ORS+∠SRT=180
∘
---Angles on a straight line
∠ORS=180
∘
−152
∘
=28
∘
y=90
∘
---Diagonals of a Rhombus are orthogonal/perpendicular to each other
In △ORS,
∠OSR+∠SOR+∠ORS=180
∘
---sum of interior angles of a triangle
∠OSR=180
∘
−90
∘
−28
∘
=62
∘
∠OSR=∠OQP=x=62
∘
---Alternate angles
∠SRO=∠OPQ=28
∘
---Alternate angles
z=∠OPQ=28
∘
---Diagonal bisects the angle at vertices, in Rhombus
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