The side `B C` of ` A B C` is produced to a point `Ddot` The bisector of `/_A` meets side `B C` in `Ldot` If `/_A B C=30^0a n d\ /_A C D=115^0,` then `/_A L C=` `85^0` (b) `72 1/2^0` (c) `145^0` (d) none of these

Explanation

`/_ACD and /_ACL make a linear pair.
/_ACD+ /_ACB = 180@
115° + ∠ACB =180°
∠ACB = 180° – 115°
∠ACB = 65°`
We know that the sum of all angles of a triangle is 180°.
Therefore, for `/_\ ABC, we have ∠ABC + ∠BAC + ∠ACB = 180°
30° + ∠BAC + 65° = 180°
∠BAC = 85°
∠LAC = ∠BAC/2 = 85/2
Using the same steps for △ALC, we get ∠ALC + ∠LAC + ∠ACL = 180°
∠ALC + 82/2 + 65°
= 180° We know that
∠ALC = ∠ACB
∠ALC = 180° – 82/2 – 65°
∠ALC = 72 ½°`