Answer:
a = 139.1
b = 56.2
Step-by-step explanation:
A. Reference angle = 68°
Opp = a
Hyp = 150
Therefore:
Sin 68 = opp/hyp
Sin 68 = a/150
150*sin 68 = a
a = 139.1 (nearest tenth)
B. Reference angle = 68°
Adj = b
Hyp = 150
Therefore:
Cos 68 = adj/hyp
Cos 68 = b/150
150*cos 68 = b
b = 56.2 (nearest tenth)
Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to

sq. units.
Step-by-step explanation:
expanded = 3n + 18
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