Answer:94.8 is the answer
Explanation:
To calculate the angle between the line BH and the plane ABCD, we first need to find the length of the line BH and the lengths of the sides of the triangle BHC.
Using the Pythagorean theorem, we can find the length of BH:
BH^2 = BG^2 + GH^2
BH^2 = 5^2 + 7^2
BH^2 = 25 + 49
BH^2 = 74
BH ≈ √74
BH ≈ 8.6 cm
To find the lengths of the sides of triangle BHC, we can use the Pythagorean theorem again:
BC^2 = BH^2 + CH^2
BC^2 = (8.6)^2 + 2^2
BC^2 = 73.96 + 4
BC^2 = 77.96
BC ≈ √77.96
BC ≈ 8.8 cm
AC^2 = AH^2 + CH^2
AC^2 = 7^2 + 2^2
AC^2 = 49 + 4
AC^2 = 53
AC ≈ √53
AC ≈ 7.3 cm
Now, we can use the law of cosines to find the angle between line BH and plane ABCD:
cos(angle) = (BC^2 + AC^2 - AB^2) / (2 * BC * AC)
cos(angle) = (8.8^2 + 7.3^2 - 34.5^2) / (2 * 8.8 * 7.3)
cos(angle) = (77.44 + 53.29 - 1190.25) / (161.92)
cos(angle) = -1059.52 / 161.92
cos(angle) ≈ -6.53
Since the angle is between 0 and 180 degrees, we can take the inverse cosine of -6.53 to find the angle:
angle ≈ acos(-6.53)
angle ≈ 94.8 degrees
Therefore, the angle between the line BH and the plane ABCD is approximately 94.8 degrees.