PQ Parellel to RS,ANGLE PAB=70,ALS=100.find angle ABC,BAC,CAQ
Attachments:
usp488:
no dear just check the difference
we don't have any
the most significant one is...
your
u have found ACB but it wasn't asked at all
no not anyone's is significant
it's just ur thought
hope u got it!
don't understand what can I do
just do what u were doing don't chat any further plz -_-
Answers
Answered by
35
Heya,
Here's ur answer:-
Let <ABC = x , <BAC = y, <ACB = z
100° = 70° + y ( alternative angles)
=> y = 100°-70° = 30°
So, <BAC = 30°
<CAq = 180° - 100° = 80° ( linear pair property)
<ABC = <BAC + <CAq = 30° + 80° = 110°
(alternative angle property)
HOPE IT HELPED (^_^)
Here's ur answer:-
Let <ABC = x , <BAC = y, <ACB = z
100° = 70° + y ( alternative angles)
=> y = 100°-70° = 30°
So, <BAC = 30°
<CAq = 180° - 100° = 80° ( linear pair property)
<ABC = <BAC + <CAq = 30° + 80° = 110°
(alternative angle property)
HOPE IT HELPED (^_^)
Answered by
20
Hello Friends...
I'm answering you.
<pAB = 70°, <ACs = 100° and Most effective thing pq ।। rs
We know that when parallel lines are bisected by the third line, in that case the opposite Angles always be the same.
So, <pAB = <ABC = 70°
Now, come to <ACB
<ACB = 180°-<ACs = 180°-100° = 80°
So, <ACB = 80°
Now find <BAC = 180°-(<ABC+<ACB) = 180°-(70°+80°) = 180°-150° = 30°
So, <BAC = 30°
Because the sum of the three angles must be 180°
And Now, find the <CAq = 180°-(<pAB+<BAC) = 180°-(70°+30) = 180°-100° = 80°
Thankyou Buddy...
HOPE it helps U :-)
I'm answering you.
<pAB = 70°, <ACs = 100° and Most effective thing pq ।। rs
We know that when parallel lines are bisected by the third line, in that case the opposite Angles always be the same.
So, <pAB = <ABC = 70°
Now, come to <ACB
<ACB = 180°-<ACs = 180°-100° = 80°
So, <ACB = 80°
Now find <BAC = 180°-(<ABC+<ACB) = 180°-(70°+80°) = 180°-150° = 30°
So, <BAC = 30°
Because the sum of the three angles must be 180°
And Now, find the <CAq = 180°-(<pAB+<BAC) = 180°-(70°+30) = 180°-100° = 80°
Thankyou Buddy...
HOPE it helps U :-)
Similar questions