Coordinate Entry - AutoCAD

1. Which relative Cartesian entry places the next vertex at the required coordinate?
   Answer: @-57,42 When working with coordinate entry in AutoCAD, you need to distinguish between absolute and relative coordinates. Absolute coordinates reference the origin $(0,0)$, while relative coordinates reference the current point. The @ symbol signals a relative entry, meaning "offset from where I am now." Here, your current point is $(72,-18)$ and your target is $(15,24)$. To find the relative offset, subtract the current point from the destination: $\Delta X = 15 - 72 = -57$ $\Delta Y = 24 - (-18) = 24 + 18 = 42$ So the correct relative entry is @-57,42, which is answer B. The negative X means you move left, and the positive Y means you move up — which makes geometric sense when shifting from a point far to the right down to one at the upper-left. A (@15,24) is the trap for students who confuse relative and absolute coordinates — those are simply the destination's absolute values, not the offset. C (-57,42) has the correct offset values but omits the @ symbol, so AutoCAD would interpret it as an absolute coordinate, placing the vertex at $(-57, 42)$ from the origin — completely wrong. D (@57,-42) reverses the signs on both components, which would move you in the exact opposite direction from the target. A reliable strategy: always compute $\Delta X$ and $\Delta Y$ explicitly by doing destination minus current position, then prefix with @. Sign errors are the most common mistake on coordinate-entry questions.
2. Which relative polar entry specifies the required segment without rounding the stated angle?
   Answer: @100<35.5 There are 60 minutes in a degree, so 30' = 30 ÷ 60 = 0.5°, and 35°30' = 35.5°. The entry is @100<35.5. A types the minutes as if they were decimal digits (35.3° is 35°18'). B rounds to the nearest whole degree, which the question rules out. D treats the 30 as seconds instead of minutes: 35 + 30 ÷ 3,600 = 35.008333…°, shown approximately as 35.0083°.
3. Which pair of relative Cartesian entries correctly creates both remaining corners?
   Answer: @60,0 followed by @0,45 When working with relative Cartesian coordinates in AutoCAD, the key idea is that each @x,y entry measures displacement from the previous point, not from the drawing origin. Every time you enter a new point, your "zero" resets to wherever you just clicked. Starting at $250,125$, the second corner needs to be $310,125$. The displacement is $310-250=60$ in X and $125-125=0$ in Y, giving @60,0. From that new position at $310,125$, the third corner must reach $310,170$. The displacement is $310-310=0$ in X and $170-125=45$ in Y, giving @0,45. That sequence — @60,0 followed by @0,45 — is exactly option D, confirming it as correct. Option A reverses both values: @0,60 moves 60 units vertically first (landing at $250,185$, not $310,125$), and @45,0 then shifts horizontally — neither corner matches. Option B uses absolute coordinates disguised as relative ones; entering @310,125 would displace an additional 310 units in X and 125 in Y from the current position, overshooting far beyond the intended corner. Option C gets the first step right with @60,0, but then abandons relative notation entirely — @310,170 again adds 310 and 170 as raw offsets rather than landing on the target point. A reliable strategy: always subtract the previous point's coordinates from the target point's coordinates to find your @Δx,Δy values. Write out the subtraction explicitly before typing — it only takes a second and prevents the most common errors on coordinate-entry questions.
4. Which approximate endpoint results from the polar entry?
   Answer: $10.000,17.321$ When AutoCAD's polar coordinate system is customized, you must account for two system variables before doing any trigonometry: ANGBASE (where 0° is measured from) and ANGDIR (which rotational direction is positive). By default, 0° points East and angles increase counterclockwise — but this question changes both rules. Here, ANGBASE = 90° means the zero-degree reference points North (straight up, the +Y axis), and ANGDIR = clockwise means angles increase in the clockwise direction. So when you enter @20<30, the angle of 30° is measured 30° clockwise from North. That places the line pointing into the upper-right quadrant, 30° from the +Y axis (equivalently, 60° from the +X axis in standard math terms). Converting to Cartesian components: $X = 20 \times \sin(30°) = 20 \times 0.5 = 10.000$ $Y = 20 \times \cos(30°) = 20 \times 0.866 = 17.321$ The endpoint is $(10.000,\ 17.321)$, confirming A is correct. B $(17.321, 10.000)$ is the trap for students who correctly identify the two component magnitudes but swap X and Y — forgetting that sine applies to X (the offset from North) and cosine applies to Y (the component along North). C $(-10.000, 17.321)$ would result from rotating counterclockwise instead of clockwise. D $(10.000, -17.321)$ ignores the ANGBASE shift entirely and places the line below the X-axis. Whenever you see ANGBASE and ANGDIR modified, redraw a quick mental compass: establish the new zero-degree direction first, then apply the rotation direction before computing components.
5. What point does AutoCAD use for this entry?
   Answer: $-43.301,25.000$ Without the @ sign, and with Dynamic Input off, AutoCAD reads 50<150 as an absolute polar coordinate measured from the UCS origin (0,0), not from the last point. X = 50 × cos 150° = −43.301 and Y = 50 × sin 150° = 25.000, so the point is (−43.301, 25.000). A (−23.301, 35.000) is what you would get from a relative entry, @50<150, which adds the same offset to the last point (20,10). B drops the negative sign on X. D (−25.000, 43.301) is the point at 120° rather than 150°, so the sizes of the X and Y values are switched.
6. Approximately where is the next point located?
   Answer: $-35.981,10.000$ When you see relative polar coordinates in AutoCAD (the @distance<angle format), you're being asked to find a new point by traveling a specific distance in a specific direction from the current point. The key steps are: resolve the polar offset into X and Y components, then add those to the current coordinates. Starting at $(-10, 25)$, the entry @30<210 means travel 30 units at 210°. Since 210° is in the third quadrant (180° + 30°), both X and Y components will be negative. Using trigonometry: $\Delta X = 30\cos(210°) = 30 \times (-0.866) \approx -25.981$ $\Delta Y = 30\sin(210°) = 30 \times (-0.500) = -15.000$ Adding to the current point: $X_{new} = -10 + (-25.981) \approx -35.981$ $Y_{new} = 25 + (-15) = 10.000$ This confirms **C) $(-35.981, 10.000)$** is correct. Choice A ($15.981, 10.000$) uses a positive X component instead of negative — a sign error that ignores which quadrant 210° falls in. Choice B ($-35.981, 40.000$) gets the X value right but adds the Y offset instead of subtracting, treating $\sin(210°)$ as positive. Choice D ($5.000, -0.981$) appears to misapply the angle entirely, possibly confusing it with a different angle or swapping sine and cosine. A reliable strategy: always sketch the angle quickly. Angles between 180° and 270° are in the third quadrant — both components are negative. Don't let a large angle number trick you into forgetting the sign.
7. What is the absolute coordinate of the final point?
   Answer: $14,-1,7$ Each @ entry is measured from the previous point. First point: 12 − 4, −8 + 10, 5 − 7 = (8, 2, −2). Final point: 8 + 6, 2 − 3, −2 + 9 = (14, −1, 7). B flips the sign of the Y value. C (8, 2, −2) is the first point, so it stops one step early. D applies only the second offset to the start point (12 + 6, −8 − 3, 5 + 9).
8. Which entry most accurately reaches the required point?
   Answer: @75<216.87 From (40, 30) to (−20, −15), ΔX = −60 and ΔY = −45, so the distance is 75. With default angle settings, the direction is 180° + arctan(45/60) = 216.869897…°, which rounds to 216.87°. D, @75<216.87, is therefore the best listed entry. The rounded angle reaches approximately (−19.99992, −15.00011), rather than the mathematically exact endpoint. A points up and right; B points up and left. With Dynamic Input off and command-line entry, C has no @ and is interpreted from the UCS origin instead of the previous point. A relative Cartesian entry, @-60,-45, would reach the point exactly.
9. What is the approximate WCS coordinate of the new endpoint?
   Answer: $134.641,70.000$ When working with User Coordinate Systems (UCS) in AutoCAD, relative coordinates entered by the user are always interpreted in the current UCS, not the WCS. Your job is to convert those UCS-relative values back into WCS coordinates. Here, the UCS is rotated $30°$ counterclockwise from the WCS. The user enters @40,0, meaning a displacement of 40 units along the UCS X-axis. To find the WCS equivalent of this displacement, apply the 2D rotation transformation: $\Delta X_{WCS} = 40\cos(30°) = 40 \times 0.866 = 34.641$ $\Delta Y_{WCS} = 40\sin(30°) = 40 \times 0.500 = 20.000$ The starting WCS point is $100, 50$, so the new endpoint is: $X = 100 + 34.641 = 134.641, \quad Y = 50 + 20.000 = 70.000$ That confirms **B) $134.641, 70.000$** as the correct answer. **A) $140.000, 50.000$** ignores the UCS rotation entirely, treating @40,0 as a pure WCS displacement — a common mistake when students forget the coordinate transformation step. **C) $134.641, 30.000$** correctly computes the X displacement but subtracts the Y component instead of adding it, as if the rotation were clockwise. **D) $34.641, 20.000$** correctly calculates the displacement vector but forgets to add it to the starting point $100, 50$. Your study tip: always think of UCS entry as a two-step process — rotate the input vector by the UCS angle, then add it to the WCS starting point. Forgetting either step is how all three wrong answers are generated.
10. What absolute Cartesian coordinate is established by the final entry?
    Answer: $90,50$ When working with AutoCAD coordinate entry, you need to distinguish between three input types: absolute coordinates (X,Y), relative Cartesian (@ΔX,ΔY), and relative polar (@distance<angle). The @ symbol always means "relative to the last point entered." Start by establishing your points step by step. The first point is entered as 125,80 — this is absolute, so your starting position is $(125, 80)$. The second entry, @-35,20, shifts the position by $-35$ in X and $+20$ in Y, giving you $125 + (-35),\ 80 + 20 = (90, 100)$. Now for the final entry: @50<270 is relative polar, meaning travel 50 units at 270°. In AutoCAD's default setup, 0° points right (east) and angles increase counterclockwise, so 270° points straight down (south). This means $\Delta X = 50\cos(270°) = 0$ and $\Delta Y = 50\sin(270°) = -50$. Adding to $(90, 100)$: $90 + 0,\ 100 + (-50) = (90, 50)$, which confirms A is correct. Choice B, $(140, 50)$, moves down 50 correctly but also adds 50 to X, as if the move had an eastward part as well. Choice C, $(90, 150)$, comes from adding $+50$ to Y rather than subtracting, confusing 270° with 90°. Choice D, $(40, 100)$, moves 50 units left instead of down, treating 270° as if it were 180°. As a study tip, always sketch the displacement direction for polar entries — 90° is up, 270° is down, 180° is left, and 0° is right. This quick mental compass prevents the most common angle-direction mistakes on the AutoCAD exam.

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