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Question
The side BC of ΔABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD = 115°, then ∠ALC = ______.
Options
85°
- \[72\frac{1}{2}^\circ\]
145°
none of these
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Solution
The side BC of ΔABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD = 115°, then ∠ALC = `bbunderline(72 1/2)` °
Explanation:
In the given problem, BC of ΔABC is produced to point D. bisectors of ∠A meet side BC at L, ∠ABC = 30° and ∠ACD = 115°
Here, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,
In ΔABC
∠ACD = ∠CAB + ∠CBA
115° = ∠CAB + 30°
∠CAB = 115° - 30°
∠CAB = 85°
Now, as AL is the bisector of ∠A
∠CAL = 1/2 ∠CAB
∠CAL = 1/2 (85°)
`∠CAL = 44 (1^\circ)/2`
Also, ∠ACD is the exterior angle of ΔALC
Thus,
Again, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,
In ΔALC
∠ACD = ∠CAL + ∠ALC
`115= 44 (1^\circ)/2+∠ALC`
`∠ALC =115- 44 1^\circ/2`
`∠ALC = 72 (1^\circ)/2`
Thus, `∠ALC = 72 1^\circ/2 `
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