Question

Calcula en cada caso las restantes razones trigonométricas de un angulo agudo si se conoce que:

Answer 1

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.

Answer 2

(A). The value of $\cos a=\dfrac{{\sqrt {22} }}{5}{\text{ and }}\tan a=\dfrac{{\sqrt {66} }}{{22}}.$

(B). The value of $\sin a=\dfrac{{2\sqrt 2 }}{3}$ and $\tan a=\dfrac{{2\sqrt 2 }}{1}.$

(C). The value of $\sin a=\dfrac{{\sqrt 5 }}{{\sqrt 6 }}$ and $\cos a=\dfrac{1}{{\sqrt 6 }}.$

(D). The value of $\sin a =\dfrac{3}{5}$ and $\tan a=\dfrac{3}{4}.$

(E). The value of $\sin a=\dfrac{5}{{\sqrt {26} }}$ and $\cos a=\dfrac{1}{{\sqrt {26}}}.$

(F). The value of $\sin a=\dfrac{6}{10}$ and $\tan a=\dfrac{6}{8}.$

Further explanation:

The Pythagorean formula can be expressed as,

$\boxed{{H^2} = {P^2} + {B^2}}.$

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

$\boxed{\sin a = \frac{P}{H}}$

The formula for cos of angle a can be expressed as

$\boxed{\cos a = \frac{B}{H}}$

The formula for tan of angle a can be expressed as

$\boxed{\tan a = \frac{P}{B}}$

Given:

(A) $\sin a =\dfrac{{\sqrt 3 }}{5}$

(B)$\tan a = \sqrt 5$

(C)$\tan a = \sqrt 5$

(D)$\cos a= \dfrac{4}{5}$

(E)$\tan a = 5$

(F)$\cos a = 0.8$

Explanation:

(A)

$\sin a= \dfrac{{\sqrt 3 }}{5}$

The perpendicular is \sqrt 3 and the hypotenuse is 5.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{5^2}&= {\left({\sqrt 3 }\right)^2} + {B^2}\\25&= 3 + {B^2}\\25 - 3&= {B^2}\\22&= {B^2}\\\end{aligned}$

The $\cos$ a can be calculated as follows,

$\cos a = \dfrac{{\sqrt {22} }}{5}$

The value of \tan a can be calculated as follows,

$\tan a= \sqrt {\dfrac{3}{{22}}}$

(B)

$\cos a= \dfrac{1}{3}$

The base is 1 and the hypotenuse is 3.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{3^2}&= {\left( 1 \right)^2} + {P^2}\\9&= 1 + {P^2}\\9 - 1&= {P^2}\\8&= {P^2}\\\sqrt8&= P\\\end{aligned}$

The $\sin$ a can be calculated as follows,

$\sin a =\dfrac{{2\sqrt 2 }}{3}$

The value of $\tan$ a can be calculated as follows,

$\tan a =\dfrac{{2\sqrt 2 }}{1}$

(C)  

$\tan a =\dfrac{{\sqrt 5 }}{1}$

The base is 1 and the perpendicular is \sqrt 5.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{H^2}&= {\left( 1 \right)^2} + {\left({\sqrt 5 }\right)^2}\\{H^2}&=1 + 5\\{H^2}&=6\\H&=\sqrt6\\\end{aligned}$

The $\sin$ a can be calculated as follows,

$\sin a =\dfrac{{\sqrt 5 }}{{\sqrt 6 }}$

The value of $\cos$ a can be calculated as follows,

$\cos a=\dfrac{1}{{\sqrt 6 }}$

(D)

$\cos a =\dfrac{4}{5}$

The base is 4 and the hypotenuse is 5.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{5^2}&= {\left(4\right)^2} + {P^2}\\25&= 16 + {P^2}\\25 - 16&= {P^2}\\9&= {P^2}\\\sqrt9&= P\\\end{aligned}$

The $\sin$ a can be calculated as follows,

$\sin a=\dfrac{3}{5}$

The value of $\tan$ a can be calculated as follows,

$\tan a =\dfrac{3}{4}$

(E)

$\tan a= \dfrac{5}{1}$

The base is 1 and the perpendicular is 5.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{H^2}&= {5^2} + {1^2}\\{H^2}&= 25 + 1\\{H^2}&=26\\H&=\sqrt {26}\\\end{aligned}$

The $\sin$ a can be calculated as follows,

$\sin a= \dfrac{5}{{\sqrt {26} }}$

The value of $\cos$ a can be calculated as follows,

$\cos a =\dfrac{1}{{\sqrt {26} }}$

(F)

$\begin{aligned}\cos a&= 0.8\\&= \frac{8}{{10}}\\\end{aligned}$

Value of $\cos a =\dfrac{8}{{10}}.$

The base is 8 and the hypotenuse is 10.

The base can be calculated with the help of Pythagorean formula.

$\begin{aligned}{10^2}&= {8^2} + {P^2}\\100&= 64 + {P^2}\\100 - 64&= {P^2}\\36&= {P^2}\\6&= P\\\end{aligned}$

The $\sin$ a can be calculated as follows,

$\sin a = \dfrac{6}{10}$

The value of \tan a can be calculated as follows,

$\tan a =\dfrac{6}{8}$

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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.