Question

The exterior ∠PRS of ∠PQR is 105°. If Q=70°, find ∠P. Is ∠PRS >∠P ?

Answer 1

IN ∆ PQR,

< Q = 70 °

< PRS = 105 °

let < QRP be angle 1,

then;

< 1+ 105 ° = 180 ° (linear pair)

< 1 = (180 - 105) °
             = 75 π

now,

we have < Q = 70 °, and
                                 < 1 = 75 ° 

let the unknown angle i.e. < P be x °

now,

< (Q + 1 + x) = 180 ° (angle sum proplerty of a triangle)
   
  (70 + 75 + x) °  = 180
 
     (145 + x )° = 180 
 
   x = (180 - 145)
 
      = 35 °

HENCE, < P = 35°

Answer 2

Hi ,

We know that ,

It in ∆PQR one side QR extended to S,

exterior angle at R = Sum of interior opposite

<PRS = <P + <Q

<PRQ > angle P

<PRQ > angle Q

[ Since Euclid's Postulate ,
whole is greater than its parts. ]

Or

<P + <Q = <PRQ

<P + 70° = 105°

<P = 105 - 70

<P = 35°

105° > 35°

<PRQ > angle P

I hope this helps you.

: )

opposite angles