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t N.
In the figure shown below
Answer:
A horizontal line segment M K intersects with line segment J L at their midpoint N.
∠J N M =(5x+2)°
∠ LN M=3( x+ 14)°
So, ∠J N M + ∠ LN M =180°[ These two angles form linear pair.Angles forming linear pair are supplementary.]
⇒5 x+ 2+ 3 (x+ 14) =180 [ By Substitution]
⇒ 5 x+2 +3 x+42°= 180°
⇒ 8 x=180°-44°
⇒8 x= 136°
⇒x= 136°÷8
⇒x=17°
So, ∠J N M =5×17 +2=87°
∠ LN M= 3×(17 +14)=3×31=93
∠J N M =∠K N L [Vertically opposite angles]
∠K N L=87°
ANSWER:
The surface area of the star is 3.2700 x square kilometres.
EXPLANATIONS:
Diameter of the star = 1.8083 x Km.
Surface area of the star = 4
Where n is the radius of the star.
So that;
n =
= 0.90415 x
n = 0.90415 x Km
Thus,
Surface area = 4 x
= 326994889
Surface area = 3.2700 x
Therefore, the surface area of the star is 3.2700 x square kilometres.