Question
How to find the altitude of an isosceles triangle with a base of 40 cm and a vertex angle of 60 degrees
Answers
GPT-4o mini
To find the altitude of an isosceles triangle with a base of 40 cm and a vertex angle of 60 degrees, we can follow these steps:
1. **Identify triangle components**: In an isosceles triangle, we can denote the two equal sides as \(a\), the base as \(b\), and the vertex angle as \(\theta\). Here, \(b = 40\) cm and \(\theta = 60^\circ\).
2. **Divide the triangle**: By drawing the altitude from the vertex to the midpoint of the base, we create two right triangles. The altitude (denote it as \(h\)) divides the base into two equal parts, therefore each part is \(b/2 = 20\) cm.
3. **Using trigonometric functions**: In one of the right triangles formed, we have:
- The angle at the top (vertex angle) is \(30^\circ\) (since it's split in half).
- The adjacent side (half the base) is \(20\) cm, and we need to find the opposite side (the altitude \(h\)).
4. **Applying the tangent function**:
\[
\tan(30^\circ) = \frac{h}{20}
\]
We can solve for \(h\):
\[
h = 20 \cdot \tan(30^\circ)
\]
5. **Knowing \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)**:
\[
h = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \approx 11.55 \text{ cm}
\]
Thus, the altitude of the isosceles triangle with a base of 40 cm and a vertex angle of 60 degrees is approximately \(11.55\) cm.
1. **Identify triangle components**: In an isosceles triangle, we can denote the two equal sides as \(a\), the base as \(b\), and the vertex angle as \(\theta\). Here, \(b = 40\) cm and \(\theta = 60^\circ\).
2. **Divide the triangle**: By drawing the altitude from the vertex to the midpoint of the base, we create two right triangles. The altitude (denote it as \(h\)) divides the base into two equal parts, therefore each part is \(b/2 = 20\) cm.
3. **Using trigonometric functions**: In one of the right triangles formed, we have:
- The angle at the top (vertex angle) is \(30^\circ\) (since it's split in half).
- The adjacent side (half the base) is \(20\) cm, and we need to find the opposite side (the altitude \(h\)).
4. **Applying the tangent function**:
\[
\tan(30^\circ) = \frac{h}{20}
\]
We can solve for \(h\):
\[
h = 20 \cdot \tan(30^\circ)
\]
5. **Knowing \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)**:
\[
h = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \approx 11.55 \text{ cm}
\]
Thus, the altitude of the isosceles triangle with a base of 40 cm and a vertex angle of 60 degrees is approximately \(11.55\) cm.