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Difficulty: Very hardPythagorean Theorem and Special Right Triangles

In the standard (x,y)(x,y) coordinate plane, a line segment OQOQ connects the origin O(0,0)O(0,0) to a point QQ in the first quadrant. The segment OQOQ has a length of 1010 units and makes an angle of 60∘60^\circ with the positive xx-axis. An isosceles right triangle △OQR\triangle OQR is constructed such that the right angle is at QQ, the leg QRQR has a length of 1010 units, and point RR lies in the first quadrant. What are the coordinates of point RR?

  1. A
    (53−5,53+5)(5\sqrt{3} - 5, 5\sqrt{3} + 5)
  2. B
    (53+5,5−53)(5\sqrt{3} + 5, 5 - 5\sqrt{3})
  3. (5+53,53−5)(5 + 5\sqrt{3}, 5\sqrt{3} - 5)Answer
  4. D
    (20−103,10)(20 - 10\sqrt{3}, 10)
  5. E
    (10+103,103−10)(10 + 10\sqrt{3}, 10\sqrt{3} - 10)

Answer

The coordinates (5+53,53−5)(5 + 5\sqrt{3}, 5\sqrt{3} - 5)
The correct answer is (5+53,53−5)(5 + 5\sqrt{3}, 5\sqrt{3} - 5) because constructing two helper 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ right triangles allows us to determine both the coordinates of QQ as (5,53)(5, 5\sqrt{3}) and the horizontal and vertical shifts to RR as +53+5\sqrt{3} and −5-5 respectively.

Step-by-Step Solution

1
Project point QQ onto the xx-axis to form a 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ right triangle.
The horizontal leg is 10cos⁡(60∘)=510 \cos(60^\circ) = 5 and the vertical leg is 10sin⁡(60∘)=5310 \sin(60^\circ) = 5\sqrt{3}.
The hypotenuse OQOQ has a length of 1010 and makes a 60∘60^\circ angle with the positive xx-axis.
2
Determine the coordinates of point QQ.
Q=(5,53)Q = (5, 5\sqrt{3}).
Point QQ is in the first quadrant, so both coordinates are positive.
3
Determine the orientation of segment QRQR.
Segment QRQR must make a 30∘30^\circ angle below the horizontal line passing through QQ (going down and to the right).
Since △OQR\triangle OQR is a right isosceles triangle with the right angle at QQ, QRQR is perpendicular to OQOQ and has length 1010. To keep RR in the first quadrant, QRQR must rotate clockwise from OQOQ by 90∘90^\circ.
4
Construct a helper 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ right triangle under QRQR to find the changes in xx and yy.
The horizontal change is +10cos⁡(30∘)=+53+10 \cos(30^\circ) = +5\sqrt{3} and the vertical change is −10sin⁡(30∘)=−5-10 \sin(30^\circ) = -5.
The hypotenuse of this triangle is QR=10QR = 10, and the angle with the horizontal is 30∘30^\circ.
5
Calculate the coordinates of RR by applying the changes to the coordinates of QQ.
R=(5+53,53−5)R = (5 + 5\sqrt{3}, 5\sqrt{3} - 5).
Add the horizontal change to xQx_Q and the vertical change to yQy_Q.

Key Concept

Solving coordinate geometry problems using 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ special right triangles.
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Topics

Pythagorean Theorem and Special Right Triangles

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