- Analyze the isosceles triangle with sides of \(10\) and vertex angle \(150^{\circ}\).
- Drop an altitude from the vertex to the base to create two congruent right triangles.
- The base angles of the isosceles triangle are calculated as \(\frac{180^{\circ} - 150^{\circ}}{2} = 15^{\circ}\).
- Using trigonometry in the resulting right triangle, the relationship between the base \(y\) and the congruent sides is \(y = 20 \cdot \cos(15^{\circ})\).
- Final Answer: \(\boxed{y = 20 \cdot \cos(15^{\circ})}\)
Explanation
- Analysis of the isosceles triangle properties
We are given an isosceles triangle with two congruent sides, \(AB = AC = 10\) inches. The vertex angle between these two sides is \(\angle BAC = 150^{\circ}\). We want to find the length of the base \(BC = y\).
To solve this, we can drop an altitude from the vertex \(A\) to the base \(BC\). Let this altitude meet the base at point \(M\). Since the triangle \(\triangle ABC\) is isosceles with \(AB=AC\), the altitude \(AM\) bisects both the vertex angle and the base \(BC\). Thus:
- \(\angle BAM = \angle CAM = \frac{150^{\circ}}{2} = 75^{\circ}\)
- \(M\) is the midpoint of \(BC\), so \(BM = MC = \frac{y}{2}\)
- \(\triangle ABM\) and \(\triangle ACM\) are right-angled triangles with the right angle at \(M\).
- Relating the base \(y\) to the angles and side lengths
In the right-angled triangle \(\triangle ABM\), we know:
- The hypotenuse \(AB = 10\) inches.
- The angle \(\angle BAM = 75^{\circ}\).
- The side opposite to \(\angle BAM\) is \(BM = \frac{y}{2}\).
Using the definition of the sine function in a right triangle, \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\), we have:
\(\sin(75^{\circ}) = \frac{BM}{AB} = \frac{y/2}{10} = \frac{y}{20}\)
Therefore, \(y = 20 \cdot \sin(75^{\circ})\).
Wait, the options provided use \(15^{\circ}\). Let's use the identity \(\sin(75^{\circ}) = \sin(90^{\circ} - 15^{\circ}) = \cos(15^{\circ})\).
So, \(y = 20 \cdot \cos(15^{\circ})\).
Alternatively, let's look at the other base angle. Since the sum of angles in \(\triangle ABC\) is \(180^{\circ}\), the two base angles are equal:
\(\angle ABC = \angle ACB = \frac{180^{\circ} - 150^{\circ}}{2} = \frac{30^{\circ}}{2} = 15^{\circ}\)
In the right-angled triangle \(\triangle ABM\):
- The hypotenuse is \(AB = 10\).
- The angle \(\angle ABM = 15^{\circ}\).
- The side opposite to \(\angle ABM\) is \(AM\).
- The side adjacent to \(\angle ABM\) is \(BM = y/2\).
Using the definition of the cosine function, \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\):
\(\cos(15^{\circ}) = \frac{BM}{AB} = \frac{y/2}{10} = \frac{y}{20}\)
Therefore, \(y = 20 \cdot \cos(15^{\circ})\).
- Conclusion
Comparing our result \(y = 20 \cdot \cos(15^{\circ})\) with the provided options:
- \(y = 20 \cdot \tan(15^{\circ})\)
- \(y = 20 \cdot \sin(150^{\circ})\)
- \(y = 20 \cdot \cos(15^{\circ})\)
- \(y = 10 \cdot \cos(15^{\circ})\)
- \(y = 20 \cdot \sin(15^{\circ})\)
- None of the above.
Our calculated expression is \(y = 20 \cdot \cos(15^{\circ})\), which is one of the given choices.
Examples
Architects and engineers often use these trigonometry principles when designing roof trusses. By treating a roof's cross-section as an isosceles triangle, they can quickly calculate the necessary span (the base) of the roof by knowing the rafter length and the pitch angle. This ensures the building structure is both stable and correctly proportioned according to the intended design.