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secondary 4 | A Maths
4 Answers Below
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Very confused on how to solve qn 7,10 and 11. Would appreciate detailed worked solution asap. Thanks!!
Date Posted: 4 years ago
Views: 426
11.
A+B+C = 180°
cos (A+B) = cos A cos B - sin A sin B
A+B = 180° - C
cos (A+B) = cos (180° - C)
= cos 180° cos C - sin 180° sin C
= (-1)cos C - (0) x sin C
= -cos C
hence
cos (A+B) = -cos C
cos A cos B - sin A sin B = -cos C
cos A cos B - sin A sin B + cos C = 0
*in case u ask like stdnts do:
how u know to start with cos (A+B) ?
from the identity
cos A cos B - sin A sin B = cos (A+B)
the rest is to follow thru'
.
.
.
A+B+C = 180°
cos (A+B) = cos A cos B - sin A sin B
A+B = 180° - C
cos (A+B) = cos (180° - C)
= cos 180° cos C - sin 180° sin C
= (-1)cos C - (0) x sin C
= -cos C
hence
cos (A+B) = -cos C
cos A cos B - sin A sin B = -cos C
cos A cos B - sin A sin B + cos C = 0
*in case u ask like stdnts do:
how u know to start with cos (A+B) ?
from the identity
cos A cos B - sin A sin B = cos (A+B)
the rest is to follow thru'
.
.
.
Thanks!!
See 4 Answers
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Good evening Cheryl! Here are my workings for Q7.
Date Posted:
4 years ago
Thank you!!!
Thanks!!
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Good evening Cheryl! Here are my workings for Q11 and Q10a.
Date Posted:
4 years ago
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Good evening Cheryl! Here are my workings for Q10b and Q10c.
Date Posted:
4 years ago
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Good evening Cheryl! Here are my workings for Q10d.
Date Posted:
4 years ago
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