The area of the triangle below is square meters. What is the length of the base?
2 answers:
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Answer:
<h2><em>46.4°</em></h2>Step-by-step explanation:
Find the diagrammatic representation in the attachment given.
We can see that the diagram is a right angled triangle. To get the angle of elevation, we will use one of the trigonometry identity SOH, CAH, TOA.
Given the adjacent of the triangle to be 2miles (observation spot from the point of launch)and the opposite (side facing the elevation angle) as 2.1miles (the height), we will use TOA.
tan∅ = opposite/adjacent
tan∅ = 2.1/2
tan∅ = 1.05
∅ = tan⁻¹1.05
∅ = 46.4°
<em>Hence the angle of elevation from Kim to the space shuttle, which is at a height of 2.1 miles is approximately 46.4°</em>
See the picture attached with letters to better understand the problem
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units