Login

And Q3 sol plz

Dear Student,

O p p o s i t e a n g l e s o f p a r r a l e \log r a m a r e e q u a l T h e r e f o r e, 6 x - 17 ° = x + 63 ° 5 x = 80 ° x = \frac{80 °}{5} = 16 ° 6 x - 17 ° = 6 \times 16 ° - 17 ° = 96 ° - 17 ° = 79 ° L e t t h e a n g l e a d j a c e n t t o 79 ° i n t h e p a r a l l e \log r a m i s y S u m o f a d j a c e n t a n g l e s o f p a r a l l e \log r a m = 180 ° 79 ° + y = 180 ° y = 180 ° - 79 ° y = 101 ° A n g l e s o f p a r a l l e \log r a m a r e 79 °, 101 °, 79 ° a n d 101 °

Regards,

0

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?