Step 1: Analyze Triangle BDC
Identify known angles in the right-angled triangle $\triangle BDC$.
- Given $BD \perp AC$, meaning $\angle BDC = 90^\circ$.
- Given $\angle DBC = 40^\circ$.
- Calculate $\angle BCD$ using the angle sum property ($180^\circ$) for $\triangle BDC$: $\angle BCD = 180^\circ - 90^\circ - 40^\circ = 50^\circ$.
- Therefore, $\angle BCA = 50^\circ$.
Step 2: Analyze Triangle AEC
Use the known angle $\angle BCA$ and the given angle $\angle CAE$ in $\triangle AEC$.
- The angle $\angle ACE$ is the same as $\angle BCA$, so $\angle ACE = 50^\circ$.
- Given $\angle CAE = 37^\circ$.
- Calculate $\angle AEC$ using the angle sum property ($180^\circ$) for $\triangle AEC$: $\angle AEC = 180^\circ - (\angle CAE + \angle ACE) = 180^\circ - (37^\circ + 50^\circ) = 180^\circ - 87^\circ = 93^\circ$.
Step 3: Calculate Angle AEB
Determine $\angle AEB$ using the relationship between $\angle AEB$ and $\angle AEC$.
- The point E lies on the line segment BC.
- Angles $\angle AEB$ and $\angle AEC$ form a linear pair along the line BC at point E. They are supplementary angles.
- Thus, $\angle AEB + \angle AEC = 180^\circ$.
- Calculate $\angle AEB$: $\angle AEB = 180^\circ - \angle AEC = 180^\circ - 93^\circ = 87^\circ$.
Final Answer
The measure of angle $\angle AEB$ is $87^\circ$.
Similar Questions
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Among the following options, which are NOT sides of a triangle?
In a Δ ABC, if ∠A = 120° and AB = AC, then the values of ∠B and ∠C are respectively:
In the equilateral Δ ABC, the base BC is trisected at D and E. The line through D, Parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. If EG and DF intersect at H, then what is the ratio of the sum of the area of parallelogram AGHF and the area of the Δ DHE to the area of the Δ ABC?
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A man goes 24 m towards east and then 10 m towards north. How far is he away from his initial position?
