In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is ta
n(b°)? triangle KL in which angle K is a right angle and angle L measures b degrees tan(b°) = five thirds tan(b°) = four thirds tan(b°) = three fourths tan(b°) = five fourths
1 answer:
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)
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The correct answer is: Answer choice: [A]:
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→ "
" ; " { u 
± 3 } " ;
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
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Explanation:
__________________________________________________________
We are asked to simplify:
;
Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;
And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
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Let us rewrite as:
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→
;
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→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
" ;
→ And we have:
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→ " " ; that is: " u / (u − 3) " ; { u
} .
and: { u
} .
→ which is: "Answer choice: [A] " .
_________________________________________________________
NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;
and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:
"u3" ;
→ Note: To solve: "u + 3 = 0" ;
Subtract "3" from each side of the equation;
→ " u + 3 − 3 = 0 − 3 " ;
→ u = -3 (when the "denominator" equals "0") ;
→ As such: " u -3 " ;
Furthermore, consider the initial (unsimplified) given expression:
→ ;
Note: The denominator is: "(u² − 9)" .
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________
→ " u² − 9 = 0 " ;
→ Add "9" to each side of the equation ;
→ u² − 9 + 9 = 0 + 9 ;
→ u² = 9 ;
Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;
→ √(u²) = √9 ;
→ | u | = 3 ;
→ " u = 3" ; AND; "u = -3 " ;
We already have: "u = -3" (a value at which the "denominator equals "0") ;
We now have "u = 3" ; as a value at which the "denominator equals "0");
→ As such: " u" ; "u -3 " ;
or, write as: " { u ± 3 } " .
_________________________________________________________
__________________________________________________________
→ "
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:
Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;
And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________
→
___________________________________________________________
→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
→ And we have:
_________________________________________________________
→ "
and: { u
→ which is: "Answer choice: [A] " .
_________________________________________________________
NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;
and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:
"u
→ Note: To solve: "u + 3 = 0" ;
Subtract "3" from each side of the equation;
→ " u + 3 − 3 = 0 − 3 " ;
→ u = -3 (when the "denominator" equals "0") ;
→ As such: " u
Furthermore, consider the initial (unsimplified) given expression:
→
Note: The denominator is: "(u² − 9)" .
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________
→ " u² − 9 = 0 " ;
→ Add "9" to each side of the equation ;
→ u² − 9 + 9 = 0 + 9 ;
→ u² = 9 ;
Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;
→ √(u²) = √9 ;
→ | u | = 3 ;
→ " u = 3" ; AND; "u = -3 " ;
We already have: "u = -3" (a value at which the "denominator equals "0") ;
We now have "u = 3" ; as a value at which the "denominator equals "0");
→ As such: " u
or, write as: " { u
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