In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.
For finding side a, you can answer this easily by the Law of Cosines. The equation is
a2=b2 +c2 -2bccosA
a2 = 11^2 + 8^2 -2(11)(8)(cos54)
a2 = 81.55
a = √81.55
a = 9
Then, we use the Law of Sines to find angles B and C. The formula would be
a/sinA = b/sinB = c/sinC
9/sin54° = 11/sinB
B = 80.4°
9/sin54° = 8/sinC
C = 45.6°
The answer would be: a ≈ 9, C ≈ 45.6, B ≈ 80.4
15% of R560 - 15% of R500
=> 0.15 × 560 – 0.15 × 500
=> 0.15 ( 560–500)
=> 0.15 × 60
=> Rs. 9
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Remove unnecessary parenthesis.
-5k 3 - 6k +1 - 6k 2 + 5k (simplify each term) -15k - 6k + 1 - 12k +5k = -28k +1
Answer: 28k + 1
