In ΔEFG, ∠E \cong≅∠G, GE = 7 and FG = 15. Find the length of EF.

 EF = 15

Step-by-step explanation:

The given description is that of an isosceles triangle. The base angles are congruent, therefore the sides opposite of those angles are also congruent.

The base angles are ∠E and ∠G and the vertex angle is ∠F.

The sides opposite to the base angles are EF and FG.

Thus, EF ≡ FG.

Since FG = 15 and FG = EF, then 15 = EF.

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