In the given figure, ∠ACE = 43° and ∠CAF = 62°; find the values of a, b and c.
Answer
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Hint: To solve this question, we need to study the figure very carefully and isolate the basic geometrical shapes. Once isolated, we will use various properties of geometrical figures to get the angles. Some properties we need to keep in mind are that the sum of the internal angles of a triangle is 180° and the sum of the internal angles of a cyclic quadrilateral is 360°. Also, that the sum of the opposite angles of a cyclic quadrilateral is equal 180°. With the help of these properties, we can find the values of a, b and c.
Complete step by step answer:
The figure given to us is
It is also given to us that ∠ACE = 43° and ∠CAF = 62°.
From the figure, we can see that ACE is a triangle and the sum of the internal angles of a triangle is 180°.
Therefore, 43° + 62° + ∠CEA = 180°
$\Rightarrow $ ∠CEA = 180° – 105°
$\Rightarrow $ ∠CEA = 75° = ∠DEA.
We can see from the figure that ABDE is a cyclic quadrilateral.
The sum of the opposite angles of the cyclic quadrilateral is equal to 180°.
Hence, ∠DEA + ∠ABD = 180°
∠ABD = a = 180° ─ 75°
a = 105°
From the figure, we can see that ABF is also a triangle and the property of sum of the angles of a triangle will apply for this triangle too.
Hence, b + a + ∠BCF = 180°
But ∠BCF = 62° and a = 105°
Therefore, b = 180° ─ 62° ─ 105°
b = 13°
∠DEA + ∠DEF = 180°, since they are linear pair of angles.
∠DEF = 180° ─ 75°
∠DEF = 105°
DEF is also a triangle.
Hence, c + ∠DEF + b = 180°
c = 180° ─ 13° ─ 105°
c = 62°
Therefore, a = 105°, b = 13° and c = 62°.
Note: The prerequisites for this question are basic properties of geometrical figures. The problem deals with cyclic quadrilaterals and triangles. Students are advised to be careful while recalling the properties as they seem very similar but the slightest difference can alter the solution completely.
Complete step by step answer:
The figure given to us is
It is also given to us that ∠ACE = 43° and ∠CAF = 62°.
From the figure, we can see that ACE is a triangle and the sum of the internal angles of a triangle is 180°.
Therefore, 43° + 62° + ∠CEA = 180°
$\Rightarrow $ ∠CEA = 180° – 105°
$\Rightarrow $ ∠CEA = 75° = ∠DEA.
We can see from the figure that ABDE is a cyclic quadrilateral.
The sum of the opposite angles of the cyclic quadrilateral is equal to 180°.
Hence, ∠DEA + ∠ABD = 180°
∠ABD = a = 180° ─ 75°
a = 105°
From the figure, we can see that ABF is also a triangle and the property of sum of the angles of a triangle will apply for this triangle too.
Hence, b + a + ∠BCF = 180°
But ∠BCF = 62° and a = 105°
Therefore, b = 180° ─ 62° ─ 105°
b = 13°
∠DEA + ∠DEF = 180°, since they are linear pair of angles.
∠DEF = 180° ─ 75°
∠DEF = 105°
DEF is also a triangle.
Hence, c + ∠DEF + b = 180°
c = 180° ─ 13° ─ 105°
c = 62°
Therefore, a = 105°, b = 13° and c = 62°.
Note: The prerequisites for this question are basic properties of geometrical figures. The problem deals with cyclic quadrilaterals and triangles. Students are advised to be careful while recalling the properties as they seem very similar but the slightest difference can alter the solution completely.
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