Respuesta :
A) cos a = (√22)/5; tan a = (√66)/22
B) sin a = (2√2)/3; tan a = 2√2
C) sin a = (√30)/6; cos a = (√6)/6
D) sin a = 3/5; tan a = 3/4
E) sin a = (5√26)/26; cos a = (√26)/26
F) sin a = 3/5; tan a = 3/4
Explanation
The ratio for sine is opposite/hypotenuse. This means the side opposite the angle is √3 and the hypotenuse is 5. Using the Pythagorean theorem to find the adjacent side,
(√3)² + A² = 5²
3+A² = 25
A² = 22
A=√22
This means that cos a = adjacent/hypotenuse = (√22)/5 and tan a = opposite/adjacent = (√3)/(√22) = (√66)/22.
B) The ratio for cosine is adjacent/hypotenuse; this means the side adjacent to the angle is 1 and the hypotenuse is 3. Using the Pythagorean theorem to find the side opposite the angle (p),
1² + p² = 3²
1+p² = 9
p² = 8
p=√8 = 2√2
This means that sin a = opposite/hypotenuse = (2√2)/3 and tan a = opposite/adjacent = (2√2)/1 = 2√2.
C) The ratio for tangent is opposite/adjacent; this means that the side opposite the angle is √5 and the side adjacent the angle is 1. Using the Pythagorean theorem to find the hypotenuse,
(√5)²+1² = H²
5+1=H²
6=H²
√6 = H
This means that sin a = opposite/hypotenuse = (√5)/(√6) = (√30)/6 and cos a = adjacent/hypotenuse = 1/(√6) = (√6)/6.
D) The ratio for cosine is adjacent/hypotenuse; this means that the side adjacent the angle is 4 and the hypotenuse is 5. Using the Pythagorean theorem to find the side opposite the angle, p:
4²+p²=5²
16+p²=25
p²=9
p=3
This means that sin a = opposite/hypotenuse = 3/5 and tan a = opposite/adjacent = 3/4.
E) The ratio for tangent is opposite/adjacent;; this means that the side opposite the angle is 5 and the side adjacent the angle is 1. Using the Pythagorean theorem to find the hypotenuse,
5²+1²=H²
25+1=H²
26=H²
√26 = H
This means that sin a = opposite/hypotenuse = 5/(√26) = (5√26)/26 and cos a = adjacent/hypotenuse = 1/(√26) = √26/26.
F) 0.8 = 8/10; The ratio for cosine is adjacent/hypotenuse. This means that the side adjacent the angle is 8 and the hypotenuse is 10. Using the Pythagorean theorem to find the side opposite the angle, p:
8²+p² = 10²
64+p² = 100
p² = 36
p=6
This means that sin a = opposite/hypotenuse = 6/10 = 3/5 and tan a = opposite/adjacent = 6/8 = 3/4.
B) sin a = (2√2)/3; tan a = 2√2
C) sin a = (√30)/6; cos a = (√6)/6
D) sin a = 3/5; tan a = 3/4
E) sin a = (5√26)/26; cos a = (√26)/26
F) sin a = 3/5; tan a = 3/4
Explanation
The ratio for sine is opposite/hypotenuse. This means the side opposite the angle is √3 and the hypotenuse is 5. Using the Pythagorean theorem to find the adjacent side,
(√3)² + A² = 5²
3+A² = 25
A² = 22
A=√22
This means that cos a = adjacent/hypotenuse = (√22)/5 and tan a = opposite/adjacent = (√3)/(√22) = (√66)/22.
B) The ratio for cosine is adjacent/hypotenuse; this means the side adjacent to the angle is 1 and the hypotenuse is 3. Using the Pythagorean theorem to find the side opposite the angle (p),
1² + p² = 3²
1+p² = 9
p² = 8
p=√8 = 2√2
This means that sin a = opposite/hypotenuse = (2√2)/3 and tan a = opposite/adjacent = (2√2)/1 = 2√2.
C) The ratio for tangent is opposite/adjacent; this means that the side opposite the angle is √5 and the side adjacent the angle is 1. Using the Pythagorean theorem to find the hypotenuse,
(√5)²+1² = H²
5+1=H²
6=H²
√6 = H
This means that sin a = opposite/hypotenuse = (√5)/(√6) = (√30)/6 and cos a = adjacent/hypotenuse = 1/(√6) = (√6)/6.
D) The ratio for cosine is adjacent/hypotenuse; this means that the side adjacent the angle is 4 and the hypotenuse is 5. Using the Pythagorean theorem to find the side opposite the angle, p:
4²+p²=5²
16+p²=25
p²=9
p=3
This means that sin a = opposite/hypotenuse = 3/5 and tan a = opposite/adjacent = 3/4.
E) The ratio for tangent is opposite/adjacent;; this means that the side opposite the angle is 5 and the side adjacent the angle is 1. Using the Pythagorean theorem to find the hypotenuse,
5²+1²=H²
25+1=H²
26=H²
√26 = H
This means that sin a = opposite/hypotenuse = 5/(√26) = (5√26)/26 and cos a = adjacent/hypotenuse = 1/(√26) = √26/26.
F) 0.8 = 8/10; The ratio for cosine is adjacent/hypotenuse. This means that the side adjacent the angle is 8 and the hypotenuse is 10. Using the Pythagorean theorem to find the side opposite the angle, p:
8²+p² = 10²
64+p² = 100
p² = 36
p=6
This means that sin a = opposite/hypotenuse = 6/10 = 3/5 and tan a = opposite/adjacent = 6/8 = 3/4.
(A). The value of [tex]\cos a=\dfrac{{\sqrt {22} }}{5}{\text{ and }}\tan a=\dfrac{{\sqrt {66} }}{{22}}.[/tex]
(B). The value of [tex]\sin a=\dfrac{{2\sqrt 2 }}{3}[/tex] and [tex]\tan a=\dfrac{{2\sqrt 2 }}{1}.[/tex]
(C). The value of [tex]\sin a=\dfrac{{\sqrt 5 }}{{\sqrt 6 }}[/tex] and [tex]\cos a=\dfrac{1}{{\sqrt 6 }}.[/tex]
(D). The value of [tex]\sin a =\dfrac{3}{5}[/tex] and [tex]\tan a=\dfrac{3}{4}.[/tex]
(E). The value of [tex]\sin a=\dfrac{5}{{\sqrt {26} }}[/tex] and [tex]\cos a=\dfrac{1}{{\sqrt {26}}}.[/tex]
(F). The value of [tex]\sin a=\dfrac{6}{10}[/tex] and [tex]\tan a=\dfrac{6}{8}.[/tex]
Further explanation:
The Pythagorean formula can be expressed as,
[tex]\boxed{{H^2} = {P^2} + {B^2}}.[/tex]
Here, H represents the hypotenuse, P represents the perpendicular and B represents the base.
The formula for sin of angle a can be expressed as
[tex]\boxed{\sin a = \frac{P}{H}}[/tex]
The formula for cos of angle a can be expressed as
[tex]\boxed{\cos a = \frac{B}{H}}[/tex]
The formula for tan of angle a can be expressed as
[tex]\boxed{\tan a = \frac{P}{B}}[/tex]
Given:
(A) [tex]\sin a =\dfrac{{\sqrt 3 }}{5}[/tex]
(B)[tex]\tan a = \sqrt 5[/tex]
(C)[tex]\tan a = \sqrt 5[/tex]
(D)[tex]\cos a= \dfrac{4}{5}[/tex]
(E)[tex]\tan a = 5[/tex]
(F)[tex]\cos a = 0.8[/tex]
Explanation:
(A)
[tex]\sin a= \dfrac{{\sqrt 3 }}{5}[/tex]
The perpendicular is \sqrt 3 and the hypotenuse is 5.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{5^2}&= {\left({\sqrt 3 }\right)^2} + {B^2}\\25&= 3 + {B^2}\\25 - 3&= {B^2}\\22&= {B^2}\\\end{aligned}[/tex]
The [tex]\cos[/tex] a can be calculated as follows,
[tex]\cos a = \dfrac{{\sqrt {22} }}{5}[/tex]
The value of \tan a can be calculated as follows,
[tex]\tan a= \sqrt {\dfrac{3}{{22}}}[/tex]
(B)
[tex]\cos a= \dfrac{1}{3}[/tex]
The base is 1 and the hypotenuse is 3.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{3^2}&= {\left( 1 \right)^2} + {P^2}\\9&= 1 + {P^2}\\9 - 1&= {P^2}\\8&= {P^2}\\\sqrt8&= P\\\end{aligned}[/tex]
The [tex]\sin[/tex] a can be calculated as follows,
[tex]\sin a =\dfrac{{2\sqrt 2 }}{3}[/tex]
The value of [tex]\tan[/tex] a can be calculated as follows,
[tex]\tan a =\dfrac{{2\sqrt 2 }}{1}[/tex]
(C)
[tex]\tan a =\dfrac{{\sqrt 5 }}{1}[/tex]
The base is 1 and the perpendicular is \sqrt 5.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{H^2}&= {\left( 1 \right)^2} + {\left({\sqrt 5 }\right)^2}\\{H^2}&=1 + 5\\{H^2}&=6\\H&=\sqrt6\\\end{aligned}[/tex]
The [tex]\sin[/tex] a can be calculated as follows,
[tex]\sin a =\dfrac{{\sqrt 5 }}{{\sqrt 6 }}[/tex]
The value of [tex]\cos[/tex] a can be calculated as follows,
[tex]\cos a=\dfrac{1}{{\sqrt 6 }}[/tex]
(D)
[tex]\cos a =\dfrac{4}{5}[/tex]
The base is 4 and the hypotenuse is 5.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{5^2}&= {\left(4\right)^2} + {P^2}\\25&= 16 + {P^2}\\25 - 16&= {P^2}\\9&= {P^2}\\\sqrt9&= P\\\end{aligned}[/tex]
The [tex]\sin[/tex] a can be calculated as follows,
[tex]\sin a=\dfrac{3}{5}[/tex]
The value of [tex]\tan[/tex] a can be calculated as follows,
[tex]\tan a =\dfrac{3}{4}[/tex]
(E)
[tex]\tan a= \dfrac{5}{1}[/tex]
The base is 1 and the perpendicular is 5.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{H^2}&= {5^2} + {1^2}\\{H^2}&= 25 + 1\\{H^2}&=26\\H&=\sqrt {26}\\\end{aligned}[/tex]
The [tex]\sin[/tex] a can be calculated as follows,
[tex]\sin a= \dfrac{5}{{\sqrt {26} }}[/tex]
The value of [tex]\cos[/tex] a can be calculated as follows,
[tex]\cos a =\dfrac{1}{{\sqrt {26} }}[/tex]
(F)
[tex]\begin{aligned}\cos a&= 0.8\\&= \frac{8}{{10}}\\\end{aligned}[/tex]
Value of [tex]\cos a =\dfrac{8}{{10}}.[/tex]
The base is 8 and the hypotenuse is 10.
The base can be calculated with the help of Pythagorean formula.
[tex]\begin{aligned}{10^2}&= {8^2} + {P^2}\\100&= 64 + {P^2}\\100 - 64&= {P^2}\\36&= {P^2}\\6&= P\\\end{aligned}[/tex]
The [tex]\sin[/tex] a can be calculated as follows,
[tex]\sin a = \dfrac{6}{10}[/tex]
The value of \tan a can be calculated as follows,
[tex]\tan a =\dfrac{6}{8}[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Trigonometry
Keywords: perpendicular bisectors, sides, right angle triangle, triangle, altitudes, hypotenuse, on the triangle, hypotenuse, trigonometric functions.
