At sunset, a surveyor standing at point O observes the shadow of a tall communication towe… — State PSC Exam

At sunset, a surveyor standing at point $O$ observes the shadow of a tall communication tower pointing directly towards point $A$ . He walks $12\text{ m}$ straight along the shadow to reach point $A$ . From point $A$ , he turns $135^\circ$ clockwise and walks $10\sqrt{2}\text{ m}$ to reach point $B$ . At point $B$ , he takes a $90^\circ$ right turn and walks $5\sqrt{2}\text{ m}$ to point $C$ . From point $C$ , after turning $135^\circ$ anti-clockwise, he walks $15\text{ m}$ to reach point $D$ . Finally, he turns to his left and walks $5\text{ m}$ to reach point $E$ . What is the shortest distance from point $O$ to point $E$ , and in which direction is point $E$ located relative to point $O$ ?

$12\text{ m}$ , EastCevap
B
$12\text{ m}$ , West
C
$13\text{ m}$ , South-East
D
$5\text{ m}$ , North

Cevap

The shortest distance between point

O

and point

E

12\text{ m}

, and point

E

is located due East of point

O

The correct answer is

12\text{ m}

, East because resolving all movement vectors on a Cartesian grid yields final coordinates

(12, 0)

relative to the starting origin

(0, 0)

. The straight-line distance is

12\text{ m}

along the positive x-axis, which corresponds to the East direction.

Adım Adım Çözüm

Determine initial orientation from shadow context.

At sunset, the sun is in the West, so shadows point East. Thus, moving along the shadow from origin

O(0,0)

means moving East.

Shadow direction defines the primary cardinal reference axis.

Calculate coordinates of point

A

and point

B

Point

A = (12, 0)

. Turning

135^\circ

clockwise from East faces South-West. Moving

10\sqrt{2}\text{ m}

South-West gives

\Delta x = -10\text{ m}, \Delta y = -10\text{ m}

, making Point

B = (12 - 10, 0 - 10) = (2, -10)

Decompose diagonal displacement into orthogonal components.

Calculate coordinates of point

C

At point

B

(facing South-West), a

90^\circ

right turn faces North-West. Moving

5\sqrt{2}\text{ m}

North-West gives

\Delta x = -5\text{ m}, \Delta y = +5\text{ m}

, making Point

C = (2 - 5, -10 + 5) = (-3, -5)

Apply relative turn to update facing direction and component movement.

Calculate coordinates of point

D

and point

E

At point

C

(facing North-West), turning

135^\circ

anti-clockwise faces East (

135^\circ - 135^\circ = 0^\circ

). Walking

15\text{ m}

East gives Point

D = (-3 + 15, -5) = (12, -5)

. Turning left (North) and walking

5\text{ m}

gives Point

E = (12, -5 + 5) = (12, 0)

Finalize vector additions to obtain final coordinates.

Compute final displacement vector relative to origin

O(0,0)

Distance =

\sqrt{(12 - 0)^2 + (0 - 0)^2} = 12\text{ m}

. Since

x = +12

and

y = 0

, the direction is East.

Calculate magnitude and direction using standard Cartesian coordinates.

Anahtar Kavram

Multi-step vector displacement with implicit shadow orientations and non-orthogonal angular rotations.

Tahmini Süre:2m 30s

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