An isosceles triangle has a vertical angle of 116 and it's base is 8 cm long calculate it's height

Answer 1

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.

1. 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.

2. 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 $\frac{8\text{ cm}}{2}=4\text{ cm}$.

3. Use trigonometry: In one of the right triangles formed, we can use the tangent function to find the height ($h$): $$\tan (B)=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{4}$$ $$\tan ({32}^{\circ })=\frac{h}{4}$$

4. Solve for height: $$h=4\cdot \tan ({32}^{\circ })$$

5. 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.