Question

I need help pretty urgently please. I don’t understand how to solve this.

Answer 1

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;

$\overline {CE}^2 = \overline {BE}^2 + \overline {BC}^2$

$\therefore \overline {CE} = \sqrt{\overline {BE}^2 + \overline {BC}^2}$

CE = √(17² + 30²) = √1189

Similarly, we have;

$\overline {CM}^2 = \overline {BM}^2 + \overline {BC}^2$

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°