To find the final position after walking 18.0 m west and 25.0 m north, we can use vector addition. The magnitude of the resultant vector is 30.8 m and the direction is -54.2°.
To find the final position after walking 18.0 m west and 25.0 m north, we can use vector addition. We can represent the two legs of the walk as vector displacements A and B. To find the sum R = A + B, we add the x-components and y-components of the vectors separately.
The x-component is -18.0 m for vector A and 0 for vector B. The y-component is 0 for vector A and 25.0 m for vector B. Adding the x-components, we get -18.0 m. Adding the y-components, we get 25.0 m.
The magnitude of the resultant vector R is the square root of the sum of the squares of the x-component and y-component, which is sqrt(
+
) = 30.8 m.
The direction of the resultant vector R can be found using the tangent function. The angle theta is given by tan(theta) = y-component / x-component. Plugging in the values, we get tan(theta) = 25.0 / -18.0. Taking the inverse tangent, we find theta = -54.2°.
Therefore, the question may be:
Given information:
18 mi from point M to point S
54 mi from point S to point L
An angle of 117∘ 117 ∘ at point S
35 mi from point L to point T
18 mi from point T to point S
Angle T measures 91∘
The figures are congruent
96 ∘angle
56 ∘ angle
Point R
Point U
41 mi