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The image shows a diagram of an isosceles triangle, where the two equal angles are 37 degrees, and the height of the triangle is marked as 12 cm. The base of the triangle is denoted by the variable \( v \), and it is divided into two equal parts by the height.

To solve for \( v \), we can use trigonometry. The height divides the triangle into two right triangles, each with:

- One angle of 37° (acute angle at the base),
- A right angle (90°) at the height,
- The height of 12 cm as the opposite side,
- Half the base, \( \frac{v}{2} \), as the adjacent side.

Using the tangent function, we have:

\[
\tan(37^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{\frac{v}{2}}
\]

Now we can solve for \( v \).

Let me calculate that for you.The length of the base \( v \) is approximately 31.85 cm.

Would you like further details on the steps or have any other questions?

Here are some related questions for practice:

1. How would the base change if the angle were increased to 45 degrees?
2. Can you calculate the area of the triangle using the base and height?
3. What would happen to the triangle’s base if the height were doubled while keeping the angles the same?
4. How can you use the sine function to solve for the base instead of tangent?
5. How do you find the perimeter of the isosceles triangle with this base?

**Tip**: In an isosceles triangle, the height always bisects the base, creating two congruent right triangles.

In the given isosceles triangle, the two base angles are 37 degrees, and the height from the vertex to the base is 12 cm. Solve for the base of the triangle.

Learn how to calculate the base of an isosceles triangle when given the height of 12 cm and base angles of 37 degrees using trigonometric principles. This problem is suitable for students in Grades 8-10.

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