Find: | (i) \[\angle GHD-\angle HDC\] |
| (ii) \[\angle BCl+\angle HAB\]. |
A)
(i) (ii) \[{{58}^{o}}\] \[{{150}^{o}}\]
B)
(i) (ii) \[{{48}^{o}}\] \[{{150}^{o}}\]
C)
(i) (ii) \[{{58}^{o}}\] \[{{180}^{o}}\]
D)
(i) (ii) \[{{48}^{o}}\] \[{{180}^{o}}\]
Correct Answer: D
Solution :
(i) \[JK||Gl\]and DH is transversal. \[\therefore \] \[\angle GHD=\angle HDK\](Alternate angles) \[\Rightarrow \] \[\angle GHD=\angle HDC+{{48}^{o}}\] \[\Rightarrow \] \[\angle GHD-\angle HDC={{48}^{o}}\] (ii) \[Gl||EF\]and BC is transversal. \[\angle BCl+\angle CBF={{180}^{o}}\] (Co-interior angles) \[\angle BCl+{{66}^{o}}={{180}^{o}}\] \[\angle BCl={{180}^{o}}-{{66}^{o}}={{114}^{o}}\] As, \[HA||CB,\]and AB is transversal \[\angle CBF=\angle HAB\] (Corresponding angles) \[\angle HAB={{66}^{o}}\] Now. \[\angle BCl+\angle HAB={{114}^{o}}+{{66}^{o}}={{180}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec
