To find the height of the isosceles triangle with a vertical angle of \( 116^\circ \) and a base of \( 8 \) cm, we can use some trigonometry.
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Divide the triangle: Draw a line from the apex (the top point of the triangle) perpendicular to the base. This line bisects the base and creates two right triangles.
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Determine the sides:
- The vertical angle \( A \) of the triangle is \( 116^\circ \).
- Each base angle \( B \) can be calculated as: \[ B = \frac{180^\circ - 116^\circ}{2} = \frac{64^\circ}{2} = 32^\circ \]
- The length of the half-base is \( \frac{8 \text{ cm}}{2} = 4 \text{ cm} \).
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Use trigonometry: In one of the right triangles formed, we can use the tangent function to find the height (\( h \)): \[ \tan(B) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{4} \] \[ \tan(32^\circ) = \frac{h}{4} \]
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Solve for height: \[ h = 4 \cdot \tan(32^\circ) \]
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Calculate using a calculator: \[ \tan(32^\circ) \approx 0.6249 \] \[ h \approx 4 \cdot 0.6249 \approx 2.4996 \text{ cm} \]
Thus, the height of the isosceles triangle is approximately 2.50 cm.