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

Answer 1

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  `