Answer:
Angle ECM is approximately 25.786°
Step-by-step explanation:
The given details are;
The horizontal base of the triangular prism ABCDEF = ABCD
AE = 17 cm = BE
M = The midpoint of AB
AB = 16 cm
BC = 30 cm
In a triangular prism, the angle EBC = 90°
Therefore;
[tex]\overline {CE}^2 = \overline {BE}^2 + \overline {BC}^2[/tex]
[tex]\therefore \overline {CE} = \sqrt{\overline {BE}^2 + \overline {BC}^2}[/tex]
CE = √(17² + 30²) = √1189
Similarly, we have;
[tex]\overline {CM}^2 = \overline {BM}^2 + \overline {BC}^2[/tex]
Where;
BM = AB/2
∴ BM = (16 cm)/2 = 8 cm
Therefore;
CM = √(8² + 30²) = 2·√241
By trigonometric ratio in the triangle formed by the points CEM, which is the right triangle ΔCEM, we have;
cos(∠ECM) = CM/EC
∴ ∠ECM = arccos(CM/EC)
Plugging in the values gives;
∠ECM = arccos((2·√(241)/√(1189)) ≈ 25.786°
