Prove AE = x (1/sin θ + 3/cos θ) in Right Triangles

Answer

The segment $AE$ is the sum of the hypotenuses of two right-angled triangles, $AC$ and $CE$. By using the definitions of sine and cosine for the given angle $\theta$, we find that $AC = \frac{3x}{\cos \theta}$ and $CE = \frac{x}{\sin \theta}$, which sums to the required expression.

Observation of the Figure

The image displays two right-angled triangles, $\triangle ABC$ and $\triangle CDE$.

• $\triangle ABC$ has a vertical side $AB = 3x$ and a right angle at $B$.
• $\triangle CDE$ has a horizontal side $DE = x$ and a right angle at $D$.
• The points $A$, $C$, and $E$ appear to lie on a straight line, forming segment $AE$.
• The angle $\theta$ is marked as $\angle BAC$.
• Since $AB \parallel CD$ and $BC \parallel DE$, the triangles are related by parallel line properties (alternate interior angles).

Proof

Given:

1. Triangles $ABC$ and $CDE$ are right-angled ($\angle ABC = 90^\circ, \angle CDE = 90^\circ$).
2. $BC \parallel DE$ and $AB \parallel CD$.
3. $DE = x$ and $AB = 3x$.
4. $\angle BAC = \theta$.

To Prove: $AE = x \left[ \frac{1}{\sin \theta} + \frac{3}{\cos \theta} \right]$

Proof Steps:

1. Calculate the hypotenuse AC In $\triangle ABC$, we use the cosine ratio for $\angle BAC = \theta$, where $AB$ is the adjacent side and $AC$ is the hypotenuse. $\cos \theta = \frac{AB}{AC} = \frac{3x}{AC}$ This formula relates the known vertical length $3x$ to the unknown segment $AC$ using the cosine of the included angle. Rearranging for $AC$: $AC = \frac{3x}{\cos \theta}$ The length of the first segment of the path is expressed in terms of $x$, $3$, and $\cos \theta$. ⚠️ This step is required on exams to establish the first part of the sum.

2. Determine $\angle DCE$ using parallel lines Since $AB \parallel CD$, the line $AE$ acts as a transversal. Therefore, $\angle BAC$ and $\angle DCE$ are corresponding angles. $\angle DCE = \angle BAC = \theta$ This equality is based on the theorem that corresponding angles are equal when a transversal intersects parallel lines.

3. Calculate the hypotenuse CE In $\triangle CDE$, we use the sine ratio for $\angle DCE = \theta$, where $DE$ is the opposite side and $CE$ is the hypotenuse. $\sin \theta = \frac{DE}{CE} = \frac{x}{CE}$ This formula relates the horizontal base $x$ to the remaining part of the segment $AE$ using the sine of the angle. Rearranging for $CE$: $CE = \frac{x}{\sin \theta}$ The length of the second segment is found by rearranging the trigonometric ratio for the hypotenuse.

4. Sum the segments to find AE The total length $AE$ is the sum of segments $AC$ and $CE$ because they are collinear. $AE = AC + CE$ Segment addition postulate allows us to combine the two previously found lengths. Substitute the expressions from Step 1 and Step 3: $AE = \frac{3x}{\cos \theta} + \frac{x}{\sin \theta}$ Factor out common term $x$: $AE = x \left[ \frac{3}{\cos \theta} + \frac{1}{\sin \theta} \right]$ Factoring demonstrates that the total length is proportional to the base variable $x$.

Final Answer

Through trigonometric substitution and parallel line properties, we conclude: $\boxed{AE = x \left[ \frac{1}{\sin \theta} + \frac{3}{\cos \theta} \right]}$

Common Mistakes

• Confusing Sine and Cosine: Students often use $\sin \theta$ for $AC$ because it is a "vertical-leaning" triangle, forgetting that $\theta$ is adjacent to $AB$, necessitating the use of $\cos \theta$.
• Angle Identification: Failing to prove that $\angle DCE = \theta$ via parallel line properties; you cannot assume the angles are the same without citing $AB \parallel CD$.