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secondary 4 | A Maths
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Hello! I need help with this question thank you!
Date Posted: 2 years ago
Views: 284
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Hope this helps!
Date Posted:
2 years ago
Thank you!
You missed out on quite a number of solutions for Q2a, arising from the missed second quadrants.
We are solving for 0° ≤ x ≤ 360° or essentially 40° ≤ 2x + 40° ≤ 760°, so we are effectively finding two complete quadrants worth of angles, so we are expecting four solutions for x.
2x + 40° = 20°, 160°, 380°, 520°, 740°, ...
We will get x = 60°, 150°, 240°, 350° within the stated intervals.
We are solving for 0° ≤ x ≤ 360° or essentially 40° ≤ 2x + 40° ≤ 760°, so we are effectively finding two complete quadrants worth of angles, so we are expecting four solutions for x.
2x + 40° = 20°, 160°, 380°, 520°, 740°, ...
We will get x = 60°, 150°, 240°, 350° within the stated intervals.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
I stand corrected! Hope these are the right answers for you!
Date Posted:
2 years ago
tysm this was very helpful :)
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