Complete the proof. Given:

CM ⊥ AB

∠3 = ∠4

Prove:

△ AMC ≅ △ BMC

Use the information provided to complete a two-column proof.

△ AMC ≅ △ BMC ⇒ proved down

Step-by-step explanation:

Let us revise the cases of congruence  

In Δ ABC

∵ CM ⊥ AB

∴ m∠AMC = m∠BMC = 90°

∴ ∠1 ≅ ∠2

In the two triangles AMC and BMC

∵ ∠1 ≅ ∠2 ⇒ proved

∵ ∠3 ≅ ∠4 ⇒ given

∵ CM is a common side in the two triangles

∴ CM ≅ CM ⇒ common side

→ Two angles and the side joining them in the 1st triangle ≅ two angles

   and the side joining them in the 2nd triangle, then use the 3rd case

   of congruency above

∴ Δ AMC ≅ Δ BMC ⇒ by using ASA postulate

△ AMC ≅ △ BMC ⇒ proved down

Step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this