find the magnitude of the vector given below. also find the measure (in degrees) of the acute angle θ formed…

1. Assume the vector has components ((x,y)):
   • Let's assume the starting - point of the vector is at the origin ((0,0)) and the end - point of the vector can be estimated from the graph. If we assume the end - point of the vector is ((6,5)) (by looking at the grid), then the vector (\vec{v}=(x,y)=(6,5)).
2. Find the magnitude of the vector:
   • The magnitude of a two - dimensional vector (\vec{v}=(x,y)) is given by the formula (|\vec{v}|=\sqrt{x^{2}+y^{2}}).
   • Substitute (x = 6) and (y = 5) into the formula:
     • (|\vec{v}|=\sqrt{6^{2}+5^{2}}=\sqrt{36 + 25}=\sqrt{61}\approx7.81).
3. Find the angle (\theta) between the vector and the (x) - axis:
   • The formula for the angle (\theta) between a two - dimensional vector (\vec{v}=(x,y)) and the positive (x) - axis is (\tan\theta=\frac{y}{x}).
   • Substitute (x = 6) and (y = 5) into the formula: (\tan\theta=\frac{5}{6}).
   • Then (\theta=\arctan(\frac{5}{6})).
   • Using a calculator, (\theta=\arctan(\frac{5}{6})\approx39.81^{\circ}).

Explanation:

Step1: Identify vector components

Assume vector (\vec{v}=(6,5))

Step2: Calculate magnitude

Use formula (|\vec{v}|=\sqrt{x^{2}+y^{2}}), so (|\vec{v}|=\sqrt{6^{2}+5^{2}}=\sqrt{61})

Step3: Calculate angle

Use formula (\tan\theta=\frac{y}{x}), so (\theta = \arctan(\frac{5}{6}))

Answer:

(a) (7.81) (b) (39.81)