tan 44 = h/ x
x = 15 ft * sin 26
so
h = x * tan 44
= (15 ft * sin 26) (tan 44 )
so the anser is D
hope this would help you
Answer:
The first transit on one side of the trench is 18 ft higher than the other side of the trench. The transit is set at 4.5 above ground, therefore the transit lens is 18' + 4.5' = 22.5' above the location on the other side of the trench. The angle of depression is the angle formed by the horizontal and the line of sight to the other side and it is given as 31°. This forms a right angle triangle with height of 22.5' and an interior angle of 31°, Then
tan( 31°) = Height / base, with the base representing the width of the trench
base = Height / tan(31°) = 22.5' / tan(31°) = 37.45' is the width of the trench
Answers:
- x = 10
- angle CAT = 126 degrees
- angle MUD = 54 degrees
=========================================================
Explanation:
∠CAT and ∠MUD are supplementary, which means the angle measures add to 180. They form a straight line.
( m∠CAT ) + ( m∠MUD ) = 180
( 11x+16 ) + ( 4x+14 ) = 180
11x+16 + 4x+14 = 180
(11x+4x) + (16+14) = 180
15x+30 = 180
15x = 180-30
15x = 150
x = 150/15
x = 10
Let's find each angle based on this x value
- m∠CAT=11x+16 = 11*10+16 = 110+16 = 126 degrees
- m∠MUD=4x+14 = 4*10+14 = 40+14 = 54 degrees
Those two angles add to 126+54 = 180 to confirm we do indeed have supplementary angles, and confirm the correct answers.
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.
