What is the best way for solving the question from class 9th ncert math of Lines and Angles chapter of Ncert of exercise 6.3of math. What is the best way for solving this question please guide me the best way for solving this question no.1 In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.
It is given the TQR is a straight line and so, the linear pairs (i.e. TQP and PQR) will add up to 180°
So, TQP +PQR = 180°
Now, putting the value of TQP = 110° we get,
PQR = 70°
Consider the ΔPQR,
Here, the side QP is extended to S and so, SPR forms the exterior angle.
Thus, SPR (SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property)
Or, PQR +PRQ = 135°
Now, putting the value of PQR = 70° we get,
PRQ = 135°-70°
Hence, PRQ = 65°
It is given the TQR is a straight line and so, the linear pairs (i.e. TQP and PQR) will add up to 180°
So, TQP +PQR = 180°
Now, putting the value of TQP = 110° we get,
PQR = 70°
Consider the ΔPQR,
Here, the side QP is extended to S and so, SPR forms the exterior angle.
Thus, SPR (SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property)
Or, PQR +PRQ = 135°
Now, putting the value of PQR = 70° we get,
PRQ = 135°-70°
Hence, PRQ = 65°