An isosceles triangle has 2 sides equal to 0.96 inches and the measure of the angle between them is 104.5 degrees. What is the measure of the third side? Please show work.

Question

Noah Atkinson · Accepted Answer

#0.96 sqrt 2 sqrt{1 - cos 104.5º}# #AB = c = AC = b = 0.96, angle A = 104.5º# Let's use "cos rule" #a^2 = b^2 + c^2 - 2bc cos A# #BC^2 = 0.96^2 + 0.96^2 - 2 * 0.96^2 cos 104.5º# #BC = 0.96 sqrt{1 + 1 -2 cos 104.5º}#

Jayden Jackson · Answer

undefined To find the measure of the third side of an isosceles triangle with two sides equal to 0.96 inches and an angle between them of 104.5 degrees, we can use the Law of Cosines. The Law of Cosines states:

$$ c^2 = a^2 + b^2 - 2ab \cos(C) $$

Where:
- $ c $ is the length of the third side (which is also equal to the other two sides in an isosceles triangle).
- $ a $ and $ b $ are the lengths of the equal sides (0.96 inches in this case).
- $ C $ is the angle between the equal sides (104.5 degrees in this case).

Plugging in the values we have:

$$ c^2 = (0.96)^2 + (0.96)^2 - 2(0.96)(0.96) \cos(104.5^\circ) $$

Now, calculate the cosine of 104.5 degrees:

$$ \cos(104.5^\circ) \approx -0.243 $$

Plug this value back into the equation:

$$ c^2 = (0.96)^2 + (0.96)^2 - 2(0.96)(0.96)(-0.243) $$

$$ c^2 = 0.9216 + 0.9216 + 0.44518336 $$

$$ c^2 = 2(0.9216) + 0.44518336 $$

$$ c^2 = 1.8432 + 0.44518336 $$

$$ c^2 = 2.28838336 $$

Now, take the square root of both sides to find $ c $, the length of the third side:

$$ c = \sqrt{2.28838336} $$

$$ c \approx 1.51 \, \text{inches} $$

So, the measure of the third side of the isosceles triangle is approximately 1.51 inches.