Question

A wooden frame is to be constructed in the form of an isosceles trapezoid, with diagonals acting as braces to strengthen the frame. The sides of the frame each measure 5.3 feet, and the longer base measure 12.7 feet. If the angles between the sides and the longer base each measure 68.4 degrees, find the length of one brace to the nearest tenth of a foot.

Answer 1

Answer:

11.8 ft

Step-by-step explanation:

Since the length of one side, l = 5.3 ft, the longer base b = 12.7 ft and the one brace, d form a triangle with angle between the longer base and side being the angle facing the brace is θ = 68.4°, we use the cosine rule to find the length of thee brace.

So d² = l² + b² -2lbcosθ

So, substituting the values of the variables into the equation, we have

d² = (5.3 ft)² + (12.7 ft)² -2(5.3)(12.7)cos68.4°

d² = 28.09 ft² + 161.29 ft² - 134.62(0.3681)

d² = 189.38 ft² - 49.56 ft²

d² = 139.82 ft²

taking square-root of both sides, we have

d = √139.82 ft²

d = 11.82 ft

d ≅ 11.8 ft to the nearest tenth of a foot.