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Foundation program for Railways

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Foundation program for Railways

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Current Affair, Maths, English, Verbal Reasoning, Non Verbal Reasoning, General Awareness, Critical Reasoning, Verbal Ability, Vedic Maths, General Knowledge

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Maths / Triangles / Exterior Angle Property
(A Brief Glimpse of ABHYAS Content - Have aLook !!!!)

## Exterior Angle Property

Exterior Angle: If the side BC of a triangle ABC is produced to form ray BD, then $\large \angle ACD$ is called an exterior angle of $\large \Delta ABC$ at C and is denoted by ext. $\large \Delta ACD.$

Theorem:  If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

 Given:  A triangle ABC. D is a point on BC produced, forming an exterior angle $\large \angle 4.$ To Prove: $\large \angle 4=\angle 1+\angle 2$                 i.e., $\large \angle ACD=\angle CAB+\angle CBA.$ Proof:  In triangle ABC, we have              $\large \angle 1+\angle 2+\angle 3=180^{\circ}$          ...........(i) Also,    $\large \angle 3+\angle 4=180^{\circ}$              [$\large \because \angle 3$ and $\large \angle 4$ forms a linear pair]  ...(ii) Find (i) and (ii), we have            $\large \angle 1+\angle 2+\angle 3=\angle 3+\angle 4$ $\large \Rightarrow\;\;\;\angle 1+\angle 2=\angle 4$ Hence,  $\large \angle 4=\angle 1+\angle 2$      i.e. $\large \angle ACD=\angle CAB+\angle CBA$

Illustration: in the figure PQ || RS. Find angle QOR.

 Solution: Extend PQ to N $\angle PNR = \angle SRN= 120^0$     [ Alternate interior angles are equal PN || RS and RO is the transversal ] $\angle PNR + \angle QNO= 180^0$  [ Linear Pair] $120^0 + \angle QNO= 180^0$ $\Rightarrow \angle QNO= 180^0-120^0=60^0$ Now, In $\bigtriangleup NOQ$, we have $\angle PQO= \angle QNO + \angle NOQ$              [ Exterior angle Property] $100^0= \angle 60^0 + \angle NOQ$ $\angle NOQ= 100^0-60^0= 40^0$

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