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Exercise 3.1 · Q2

Q.For each given angle, find a coterminal angle θ\theta such that 0∘≤θ<360∘0^\circ \le \theta < 360^\circ:

(i) 395∘395^\circ
(ii) 525∘525^\circ
(iii) 1150∘1150^\circ
(iv) −270∘-270^\circ
(v) −450∘-450^\circ
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Since coterminal angles differ by whole multiples of 360∘360^\circ, subtract 360∘360^\circ from an angle greater than 360∘360^\circ (as many times as needed) or add 360∘360^\circ to a negative angle, until the result lies in [0∘,360∘)[0^\circ,360^\circ).

Step 1. (i) 395∘395^\circ. 395∘−360∘=35∘395^\circ-360^\circ=35^\circ, inside [0∘,360∘)[0^\circ,360^\circ).

Step 2. (ii) 525∘525^\circ. 525∘−360∘=165∘525^\circ-360^\circ=165^\circ, inside range.

Step 3. (iii) 1150∘1150^\circ. Subtract 360∘360^\circ repeatedly: 1150∘−360∘=790∘1150^\circ-360^\circ=790^\circ; 790∘−360∘=430∘790^\circ-360^\circ=430^\circ; 430∘−360∘=70∘430^\circ-360^\circ=70^\circ (three subtractions total, since 1150∘−3(360∘)=1150∘−1080∘=70∘1150^\circ-3(360^\circ)=1150^\circ-1080^\circ=70^\circ).

Step 4. (iv) −270∘-270^\circ. −270∘+360∘=90∘-270^\circ+360^\circ=90^\circ, inside range.

Step 5. (v) −450∘-450^\circ. One addition isn't enough: −450∘+360∘=−90∘-450^\circ+360^\circ=-90^\circ, still negative, so add 360∘360^\circ again: −90∘+360∘=270∘-90^\circ+360^\circ=270^\circ (equivalently −450∘+2(360∘)=270∘-450^\circ+2(360^\circ)=270^\circ).

✓Final answer

(i) 35∘35^\circ (ii) 165∘165^\circ (iii) 70∘70^\circ (iv) 90∘90^\circ (v) 270∘270^\circ

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