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
Answers:measure angle x = 40°
measure angle y = 35°
measure angle z = 55°
Explanation:Part (a): getting angle x:In triangle BED, we have:
measure angle BED = 90°
measure angle BDE = 50°
Therefore:
measure angle DBE = 180 - (90+50) = 40°
Now, we have angle DBE and angle GBF vertically opposite angles.
This means that they are both equal. Therefore angle GBF = 40°
Since angle GBF is x, therefore:
x = 40°
Part (b): getting angle y:We know that the sum of measures of angles on a straight line is 180.
This means that:
angle GBF + angle GBC + angle CBE = 180
We have:
angle GBF = 40°
angle GBC = 105°
angle CBE = y
Therefore:
40 + 105 + y = 180
y = 35°
Part (c): getting angle z:In triangle BCE, we have:
measure angle BCE = z
measure angle BEC = 90°
measure angle CBE = 35°
Therefore:
z + 90 + 35 = 180
z = 55°
Hope this helps :)
The answer is C.
All of the other answers are incorrect
Okay what problem is it that you need help with??