In △ABC,BD⊥AC at D and ∠DBC=$$44^∘$$​. E is a point on BC such that ∠CAE=$$34^∘$$​. What is the measure of ∠AEB ?

Correct option is B

Given:

In $$\triangle ABC$$​ $$BD \perp AC$$​ at D

$$\angle DBC = 44^o$$​

Formula Used:

Sum of a triangle = $$180^o$$​​

Solution:

Let the line BD and AE intersect at O as shown in the figure

In $$\triangle AOD$$​​

​$$\angle CAE = 34^o \text{ and }\angle BDA = 90^o$$​​

Then $$\angle AOD = 180^o - ( 90^o + 34^o)$$​   

​$$\angle AOD= 56^o$$​​

​$$\angle BOE = \angle AOD = 56^o$$​ ( Vertically opposite angles )

In $$\triangle BOE$$​​

​$$\angle DBC = 44^o$$​​

​$$\angle BOE = 56^o$$​​

​$$\angle AEB = 180^o - ( 44^o + 56^o)$$​​

​$$= 180^o - 100^o$$​​

​$$= 80^o$$​​