In, since AB = AC, the angles opposite to the equal sides are equal, which means =. The sum of all interior angles of a triangle is 180^°. Subtracting = 80^° leaves 100^°, so each base angle measures 50^°.

In, since AB = AC, the angles opposite to the equal sides are equal, which means =. The sum of all interior angles of a triangle is 180^°. Subtracting = 80^° leaves 100^°, so each base angle measures 50^°.

Since segments BD and CD are the angle bisectors of and, they divide each angle into 2 equal halves. Therefore, = 50^° / 2 = 25^° and = 50^° / 2 = 25^°.

In, the sum of the interior angles is 180^°. Adding and gives 25^° + 25^° = 50^°. Subtracting 50^° from 180^° yields = 130^°. Alternatively, using the incenter formula: = 90^° + ()/(2) = 90^° + 40^° = 130^°.

In, the sum of the interior angles is 180^°. Adding and gives 25^° + 25^° = 50^°. Subtracting 50^° from 180^° yields = 130^°. Alternatively, using the incenter formula: = 90^° + ()/(2) = 90^° + 40^° = 130^°.

5. एक समद्विबाहु त्रिभुज $\Delta\text{ABC}$ में, $\text{AB} = \text{AC}$ और $\angle\text{A} = 80^\circ$ तथा $\angle\text{B}$ और $\angle\text{C}$ के द्विभाजक $\text{D}$ पर मिलते हैं। $\angle\text{BDC}$ किसके बराबर है? In an isosceles triangle $\Delta\text{ABC}$ , $\text{AB} = \text{AC}$ and $\angle\text{A} = 80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at $\text{D}$ , then $\angle\text{BDC}$ is equal to? (SSC CGL Tier - I 2016) (a) $90^\circ$ (b) $100^\circ$ (c) $130^\circ$ (d) $80^\circ$ 6. दी गई आकृति में, $\angle\text{QRN} = 40^\circ$ , $\angle\text{PQR} = 46^\circ$ तथा $\text{MN}$ बिन्दु $\text{R}$ पर एक स्पर्श रेखा है। क्रमशः $x$ , $y$ तथा $z$ का मान (डिग्री में) क्या है? In the given figure, $\angle\text{QRN} = 40^\circ$ , $\angle\text{PQR} = 46^\circ$ and $\text{MN}$ is a tangent to the point $\text{R}$ . What is the value (in degrees) of $x$ , $y$ and $z$ respectively? DIAGRAM DETAILS: A circle contains an inscribed triangle $\text{PQR}$ . A tangent line $\text{MN}$ touches the circle at point $\text{R}$ . - Vertices $\text{P}$ , $\text{Q}$ , and $\text{R}$ lie on the circle forming $\triangle\text{PQR}$ . - The line $\text{MN}$ passes through $\text{R}$ , with $\text{M}$ on one side and $\text{N}$ on the other. - The angle between the tangent segment $\text{RN}$ and the chord $\text{RQ}$ , labeled $\angle\text{QRN}$ , is indicated as $40^\circ$ (or related to angle markings). - $\angle\text{PQR} = 46^\circ$ . - Angles $x$ , $y$ , and $z$ are indicated at the vertices/intersections involving the triangle and the tangent line.

Answer: (c) $130^\circ$

Step-by-step solution

Step 1: Find the base angles of isosceles triangle ABC

In $\Delta\text{ABC}$ , since $\text{AB} = \text{AC}$ , the angles opposite to the equal sides are equal, which means $\angle\text{B} = \angle\text{C}$ . The sum of all interior angles of a triangle is $180^\circ$ . Subtracting $\angle\text{A} = 80^\circ$ leaves $100^\circ$ , so each base angle measures $50^\circ$ .

Step 2: Calculate the halved angles formed by the angle bisectors

Since segments $\text{BD}$ and $\text{CD}$ are the angle bisectors of $\angle\text{B}$ and $\angle\text{C}$ , they divide each angle into 2 equal halves. Therefore, $\angle\text{DBC} = 50^\circ / 2 = 25^\circ$ and $\angle\text{DCB} = 50^\circ / 2 = 25^\circ$ .

Step 3: Determine angle BDC in triangle BDC

In $\Delta\text{BDC}$ , the sum of the interior angles is $180^\circ$ . Adding $\angle\text{DBC}$ and $\angle\text{DCB}$ gives $25^\circ + 25^\circ = 50^\circ$ . Subtracting $50^\circ$ from $180^\circ$ yields $\angle\text{BDC} = 130^\circ$ . Alternatively, using the incenter formula: $\angle\text{BDC} = 90^\circ + \frac{\angle\text{A}}{2} = 90^\circ + 40^\circ = 130^\circ$ .

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