13) 135° 14) 143°
Solution:
13) The lengths of tangents drawn from an external point are equal.
So, PA = PB
Tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
OB ⊥ PB and OA ⊥ PA
So, ∠PBO = 90°, ∠PAO = 90°
Sum of the angles in a quadrilateral is 360°.
In quadrilateral PAOB,
⇒ ∠APB + ∠PAO + ∠PBO + ∠AOB = 360°
⇒ 45° + 90° + 90° + ∠AOB = 360°
⇒ ∠AOB = 360° – 225°
⇒ ∠AOB = 135°
Hence the missing angle is 135°
14) Angle in a semi-circle is 90°
.
∠ACB = 90°
Sum of the angles in triangle is 180°
.
In ∆ABC, ∠ACB + ∠ABC + ∠CAB = 180°
⇒ 90° + 53° + ∠CAB = 180°
⇒ ∠CAB = 180° – 143°
⇒ ∠CAB = 37°
AD and DE are radius.
So, ∠DEA = 37°
Sum of the adjacent angles in a straight line is 180°.
⇒ ∠DEA + ∠DEB = 180°
⇒ 37° + ∠DEB = 180°
⇒ ∠DEB = 143°
Hence the missing angle is 143°
