To determine the size of the two equal base angles in an isosceles triangle with equal sides of 66.9 inches and a base of 92.3 inches, follow these steps:
1. Identify the components of the isosceles triangle:
- Equal sides (each): [tex]\( 66.9 \)[/tex] inches
- Base: [tex]\( 92.3 \)[/tex] inches
2. Divide the isosceles triangle into two right-angled triangles by drawing a height from the vertex opposite the base to the midpoint of the base:
- This height splits the base into two equal segments, each half the length of the base.
- Each half of the base: [tex]\( \frac{92.3}{2} = 46.15 \)[/tex] inches
3. Determine the length of the height using the Pythagorean theorem:
- Consider one of the right-angled triangles:
- Hypotenuse (equal side of the isosceles triangle): [tex]\( 66.9 \)[/tex] inches
- One leg (half of the base): [tex]\( 46.15 \)[/tex] inches
- Height (other leg): Let’s denote it as [tex]\( h \)[/tex]
- Use the Pythagorean theorem to find [tex]\( h \)[/tex]:
[tex]\[
h = \sqrt{(66.9)^2 - (46.15)^2}
\][/tex]
4. Calculate the base angle using trigonometry:
- We can use the tangent function in the right-angled triangle:
[tex]\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{46.15}
\][/tex]
- Solve for [tex]\( \theta \)[/tex]:
[tex]\[
\theta = \tan^{-1}\left(\frac{h}{46.15}\right)
\][/tex]
5. Convert the angle from radians to degrees:
6. Round the result to the nearest tenth of a degree:
Following these steps provides the result for the base angle:
[tex]\[
\theta = 46.4^{\circ}
\][/tex]
Thus, the size of the two equal base angles of the isosceles triangle is [tex]\( 46.4^{\circ} \)[/tex], and the correct answer is:
[tex]\[
\boxed{46.4^{\circ}}
\][/tex]
Therefore, the answer is:
E. [tex]\( 46.4^{\circ} \)[/tex]
