We will discuss about some of the properties of angles of a triangle.
1. The three angles of a triangle are together equal to two right angles.
ABC is a triangle.
Then ∠ZXY + ∠XYZ + ∠YZX = 180°
Using this property, let us solve some of the examples.
Solved examples:
(i) In ∆XYZ, ∠X = 55° and ∠Y = 75°. Find ∠Z.
Solution:
∠X + ∠Y + ∠Z = 180°
or, 55° + 75° + ∠Z = 180°
or, 130° + ∠Z = 180°
or, 130° - 130° + ∠Z = 180° - 130°
Therefore, ∠Z = 50°
(ii) In the ∆XYZ, ∠Y = 5∠Z and ∠X= 3∠Z. Find the angles of the triangle.
Solution:
∠X + ∠Y + ∠Z = 180°
or, 3∠Z + 5∠Z + ∠Z = 180°
or, 9∠Z = 180°
or, \(\frac{9∠Z}{9}\) = \(\frac{180°}{9}\)
Therefore, ∠Z = 20°
We know, ∠X= 3∠Z
Now, plug-in the value of ∠Z
∠X= 3 × 20°
Therefore, ∠X= 60°
Again we know, ∠Y= 5∠Z
Now, plug-in the value of ∠Z
∠Y= 5 × 20°
Therefore, ∠Y= 100°
Hence, the angles of the triangle are ∠X = 60°, ∠Y = 100° and ∠Z = 20°.
2. If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
The side QR of the ∆PQR is produced to S.
Then ∠PRS = ∠RPQ + ∠PQR
Corollary 1: An exterior angle of a triangle is greater than either of the interior opposite angles.
In ∆PQR, QR is produced to S.
Therefore, ∠PRS > ∠RPQ and ∠PRS ∠PQR
Corollary 2: A triangle can have only one right angle.
Corollary 3: A triangle can have only one obtuse angle.
Corollary 4: A triangle must have at least two acute angles.
Corollary 5: In a right-angled triangle, the acute angles are complementary.
Now, using this property, let us solve some of the following examples.
Solved examples:
(i) Find ∠Q from the given figure.
Solution:
∠P + ∠Q = ∠PRS
Given, ∠P = 50° and ∠PRS = 120°
or, 50° + ∠Q = 120°
or, 50° - 50° + ∠Q = 120° - 50°
or, ∠Q = 120° - 50°
Therefore, ∠Q = 70°
(ii) From the given figure find all the angles of ∆ABC, given that ∠B = ∠C.
Solution:
Given, ∠B = ∠C
We know, ∠DAC = 150°
∠DAC + ∠CAB = 180°, as they form a linear pair
or, 150° + ∠CAB = 180°
or, 150° - 150° + ∠CAB = 180° - 150°
or, ∠CAB = 30°
Let ∠B = ∠C = x°
Therefore, x° + x° = 150°, as the exterior angle of a triangle is equal to the sum of the interior opposite angles.
or, 2x° = 150°
or, \(\frac{2x°}{2}\) = \(\frac{150°}{2}\)
or, x° = 75°
Therefore, ∠B = ∠C = 75°.
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