A sphere with a radius of 11 in. Answer in terms of π
1 answer:
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Answer:
Tan(b°) = 3/4 = three fourths (C)
Step-by-step explanation:
Triangle JKL is a right angled triangle in which angle K is a right angle and angle L= b°
We would apply SOHCAHTOA from trigonometry to determine the value for each sides.
For ∆JKL
Sin(b°) = opposite/hypotenuse
Sin(b°) = 3/5
Cos(b°) = adjacent/hypotenuse
Cos(b°) = 4/5
Tan(b°) = 3/4
From the above we know the value of each sides of ∆JKL
Find attached the diagram of the above triangle (1)
∆JKL is dilated by a scale factor = 2
Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:
Opposite = 2(3) = 6
Adjacent = 2(4) = 8
Hypotenuse = 2(5) = 10
To find tan(b°) of the new triangle, we would apply tangent ratio
Tan(b°) = opposite/adjacent
Tan(b°) = 6/8
Tan(b°) = 3/4
Find attached the diagram for the new triangle (2)
Diagram 3 shows the drawing of both triangles together.
From the above, we can see the angle doesn't change when a shape is dilated by a scale factor.
Therefore, tan(b°) = three fourths (C)
