Analyze Triangle Geometry Problem
We are given a triangle $\triangle ABC$. A line segment $BD$ is perpendicular to $AC$ at point D ($BD \perp AC$). We know the angle $\angle DBC = 71^\circ$. Another point E lies on the side BC, and we are given $\angle CAE = 17^\circ$. The goal is to find the measure of the angle $\angle AEB$.
Calculate Angle BCA
Consider the triangle $\triangle BDC$. Since $BD \perp AC$, the angle $\angle BDC$ is a right angle ($90^\circ$). The sum of angles in any triangle is $180^\circ$. In $\triangle BDC$, we have:
- $\angle BDC = 90^\circ$
- $\angle DBC = 71^\circ$
Using the angle sum property:
$ \angle BCA = 180^\circ - \angle BDC - \angle DBC $
$ \angle BCA = 180^\circ - 90^\circ - 71^\circ $
$ \angle BCA = 19^\circ $
Since point E lies on BC, $\angle ACE$ is the same angle, so $\angle ACE = 19^\circ$.
Determine Angle AEB using Triangle ABE
Let's denote the angle $\angle BAE$ as $\alpha$.
The angle $\angle BAC$ is the sum of $\angle BAE$ and $\angle CAE$:
$ \angle BAC = \angle BAE + \angle CAE $
$ \angle BAC = \alpha + 17^\circ $
Now, consider the larger triangle $\triangle ABC$. The sum of its angles must be $180^\circ$. We can find $\angle ABC$:
$ \angle ABC = 180^\circ - \angle BAC - \angle BCA $
Substitute the expressions for $\angle BAC$ and the value of $\angle BCA$:
$ \angle ABC = 180^\circ - (\alpha + 17^\circ) - 19^\circ $
$ \angle ABC = 180^\circ - \alpha - 17^\circ - 19^\circ $
$ \angle ABC = 180^\circ - \alpha - 36^\circ $
$ \angle ABC = 144^\circ - \alpha $
Since E is on BC, $\angle ABE$ is the same as $\angle ABC$. So, $\angle ABE = 144^\circ - \alpha$.
Finally, apply the angle sum property to triangle $\triangle ABE$:
$ \angle BAE + \angle ABE + \angle AEB = 180^\circ $
Substitute the known values and expressions:
$ \alpha + (144^\circ - \alpha) + \angle AEB = 180^\circ $
The $\alpha$ terms cancel out:
$ 144^\circ + \angle AEB = 180^\circ $
Solve for $\angle AEB$:
$ \angle AEB = 180^\circ - 144^\circ $
$ \angle AEB = 36^\circ $
Conclusion
The measure of angle $\angle AEB$ is $36^\circ$.
Similar Questions
- In $\triangle ABC$, $BD \perp AC$ at $D$ and $\angle DBC = 40^\circ$. $E$ is a point on $BC$ such that $\angle CAE = 37^\circ$. What is the measure of $\angle AEB$?
- In triangle ABC, bisector of $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at O. If $\angle\text{BAC} = 60^\circ$, then find the measure of $\angle\text{BOC}$.
- In $\Delta\text{XYZ}$, $\text{XY} = 12\text{ cm}$ and $\text{YZ} = 18\text{ cm}$. $\text{XW}$, the angle bisector of $\text{YXZ}$, meets $\text{YZ}$ at $\text{W}$, such that $\text{YW} : \text{WZ}$ is $4 : 5$. Find the length of the third side of the triangle.
- In $\Delta\text{PQR}$, $\text{QR}$ is extended up to $\text{S}$ so that $\text{RS} = \text{RP}$. If $\angle\text{PRQ} = 70^\circ$ and $\angle\text{QPS} = 110^\circ$ then find the measure of $\angle\text{PQS}$.
- The areas of two similar triangles are respectively $16\text{ m}^{2}$ and $36\text{ m}^{2}$. Find the ratio of their corresponding sides.
- In $\Delta PQR$, QR is extended up to S, so that RS = RP. If $\angle RPQ = 55^\circ$ and $\angle PRS = 110^\circ$, then find the measure of $\angle PQS$.
- The sides of a triangle are 44 cm, 33 cm, and 55 cm. What is its area? (in $cm^2$)
- The perimeter of an equilateral triangle ABC is 22.2 cm. What is the area of triangle (in $cm^2$)?
- In $\triangle\text{ABC}$, $\text{BD} \perp \text{AC}$ at $\text{D}$ and $\angle\text{DBC} = 60^{\circ}$. $\text{E}$ is a point on $\text{BC}$ such that $\angle\text{CAE} = 20^{\circ}$. What is the measure of $\angle\text{AEB}$?
- In a triangle, if angle $A = 30^\circ$ and angle $B = 45^\circ$, then what is the angle of C?
Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
