WebTrig Flash Cards AcroT E X eDucation Bundle Begin FS Close Home THE UNIVERSITY OF AKRON Theoretical and Applied Mathematics Flash Cards Right triangle trigonometry Katie Jones and Tom Price Instructions: Click on the Begin button to view the first randomly selected card. Click on FS to view the cards in full screen mode (works only outside a web browser). The Home button on the first page goes to the WebTrig home page; otherwise, the Home button returns to this page. The Close button closes the doc- ument (use outside a web browser). c 2003 [email protected] Last Revision Date: April 12, 2003 Version 1.0
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THE UNIVERSITY OF AKRONTheoretical and Applied Mathematics
Flash CardsRight triangle trigonometry
Katie Jonesand
Tom Price
Instructions: Click on the Begin button to view thefirst randomly selected card. Click on FS to view thecards in full screen mode (works only outside a webbrowser). The Home button on the first page goes tothe WebTrig home page; otherwise, the Home buttonreturns to this page. The Close button closes the doc-ument (use outside a web browser).
c© 2003 [email protected] Revision Date: April 12, 2003 Version 1.0
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A right triangle contains a 50◦ an-gle that has an adjacent side oflength 6.3 units. Find the lengthof the opposite side and the hy-potenuse.
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A right triangle contains an 18◦
angle that has an opposite side oflength 1.7 units. Find the lengthof the adjacent side and the hy-potenuse.
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A right triangle contains a 62◦
angle that has a hypotenuse oflength 8.3 units. Find the lengthof the adjacent side and the op-posite side.
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Suppose sin α =611
. Withoutusing a calculator find cos α andtan α.
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Suppose cos α =35. Without
using a calculator find sin α andtan α.
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Suppose tan α =512
. Withoutusing a calculator find sin α andcos α.
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Let α denote a non-right angle ofa right triangle. If the side ad-jacent α is 9 units long and thehypotenuse is 18 units, find sin α.
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Let α denote a non-right angle ofa right triangle. If the side adja-cent α is 12.2 units long and thehypotenuse is 30 units, find cot α.
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Let α denote a non-right angle ofa right triangle. If the side op-posite α is 12 units long and thehypotenuse is 24 units, find cos α.
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Let α denote a non-right angle ofa right triangle. If the side ad-jacent α is 8 units long and thehypotenuse is 21 units, find csc α.
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Let α denote a non-right angle ofa right triangle. If the side op-posite α is 15 units long and thehypotenuse is 31 units, find sec α.
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Let α denote a non-right angleof a right triangle. If the sideadjacent α is 4.6 units long andthe hypotenuse is 9.2 units, findtan α.
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Let T be the triangle depicted inthe figure. Find sin α.
c
10 9
67α
γ
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Let T be the triangle depicted inthe figure. Find sin γ.
b
10
14
72
γ
β
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Let T be depicted in the figurebelow. Find c.
a7 75
c26α
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Let T be the triangle in the fig-ure. Find a.
a19.4
112c
27
γ
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Let T be the triangle in the figurebelow. Find cos β.
1811
9
γ
βα
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Let T be the triangle depicted inthe figure below. Find cos γ.
4542
70
γ
βα
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Let T be the triangle given in thefigure below. Find a.
14
2447
aγ
β
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Determine b in the figure.
b
1.792
2
γ
α
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HINT
To find the length of the opposite side andthe hypotenuse of a right triangle with a 50◦
angle that has an adjacent side of length 6.3units, use the triangular definition of the co-sine of an angle which is the length of theadjacent side divided by the hypotenuse.
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Answer: hyp = 9.8 and opp = 7.51
Solution: First, find thehypotenuse using the cosineratio:
cos 50◦ =6.3hyp
.
So,
hyp =6.3
cos 50◦
= 9.801 1
Next, the Pythagorean theo-rem1 suggests that
(9.801 1)2 = (6.3)2 + (opp)2 .
So,
(opp)2 = 56.372
Hence,
opp =√
56.372= 7.5081
�
1(hyp)2 = (adj)2 + (opp)2
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HINT
To find the length of the adjacent side andthe hypotenuse of a right triangle with an18◦ angle that has an adjacent side of length1.7 units, use the triangular definition of thesine of an angle which is the length of theopposite side divided by the hypotenuse.
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Answer: hyp = 5.5 and adj = 5.23
Solution: Find the hy-potenuse using the sine ratio:
sin 18◦ =1.7hyp
so,
hyp =1.7
sin 18◦
= 5.501 3
Next, the Pythagorean theo-
rem2 suggests that
(5. 501 3)2 = (adj)2 + (1.7)2 .
So,
(adj)2 = 27. 374
Hence,
adj =√
27. 374= 5. 232
�
2(hyp)2 = (adj)2 + (opp)2
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HINT
To find the length of the adjacent side andthe opposite side of a right triangle with a62◦ angle that has an adjacent side of length8.3 units, use the triangular definition of thesine of an angle which is the length of theopposite side divided by the hypotenuse.
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Answer: opp = 7.33 and adj = 3.9
Solution: Using the sineratio we get
sin 62◦ =opp8.3
.
So,
opp = 8.3 · sin 62◦
= 7.328 5
Next, the Pythagorean theo-rem3 suggests that
(8.3)2 = (adj)2 + (7. 328 5)2
so,
(adj)2 = 15. 183.
Hence,
adj =√
15. 183 = 3. 896 5
�
3(hyp)2 = (adj)2 + (opp)2
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HINT
To find cos α and tan α when
sin α =611
use the triangular definition of the sine of anangle which is sin α = opp
hyp .
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Answer:
cos α =√
8 511
and tan α =6√
8585
Solution: Given sinα =611
, use the Pythagorean the-
orem4 to get
(11)2 = (adj)2 + (6)2 .
So, (adj)2 = 85,or
adj =√
85.
Hence,
cos α =adjhyp
=√
8 511
and
tanα =oppadj
=6√85
=6√
8585
.
�
4(hyp)2 = (adj)2 + (opp)2
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HINT
To find sin α and tan α when
cos α =35,
use the triangular definition
cos α =adjhyp
.
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Answer: sin α =45
and tan α =43
Solution: Given
cos α =35
=adjhyp
,
and using the Pythagoreantheorem5 we get
(5)2 = (3)2 + (opp)2
Consequently,
opp =√
16 = 4
suggesting that
sin α =opphyp
=45
and
tanα =oppadj
=43.
�
5(hyp)2 = (adj)2 + (opp)2
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HINT
To find sin α and cos α when
tan α =512
recall the triangular definitions of these func-tions.
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Answer: sin α =513
and tan α =1213
Solution: Recall that
tanα =512
=oppadj
.
Next, find the hypotenuse us-ing the Pythagorean theo-rem6
(hyp)2 = (12)2 + (5)2
= 169.
Consequently,
opp =√
169 = 13
so that
sin α =opphyp
=513
and
cos α =adjhyp
=1213
.
�6(hyp)2 = (adj)2 + (opp)2
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HINT
To find sin α if the side adjacent α is 9 unitslong and the hypotenuse is 18 units remem-ber the triangular definition of the sine of anangle uses the length of the side opposite αand the length of the hypotenuse.
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Answer: sin α = .87Solution: Find the opposite side using the pythagorean
theorem (hyp)2 = (adj)2 + (opp)2
(18)2 = (9)2 + (opp)2
=⇒ (opp)2 = 243
=⇒ opp =√
243 = 15. 588
Then we know,
sin α =opphyp
=√
24318
= 0.866 03.
�
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HINT
To find cot α if the side adjacent α is 12.2units long and the hypotenuse is 30 units re-call that
cot α =adjopp
.
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Answer: cot α = .45Solution: Find the opposite side using the pythagorean
theorem (hyp)2 = (adj)2 + (opp)2
(30)2 = (12.2)2 + (opp)2
=⇒ (opp)2 = 751. 16
=⇒ opp =√
751. 16 = 27. 407
We knowcot α =
adjopp
=12.2
27.407= 0.445 14
�
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HINT
To find cos α if the side opposite α is 12 unitslong and the hypotenuse is 24 units use thetriangular definition of the cosine of an anglewhich uses the length of the hypotenuse andside adjacent to α.
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Answer: cos α = .87Solution: Find the adjacent side using the pythagorean
theorem (hyp)2 = (adj)2 + (opp)2
(24)2 = (adj)2 + (12)2
=⇒ (adj)2 = 432
=⇒ adj =√
432 = 20. 785
We know,
cos α =adjhyp
=√
43224
=√
32
= 0.866 03
�
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HINT
To find csc α if the side adjacent α is 8 unitslong and the hypotenuse is 21 units use
csc α =hypopp
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Answer: csc α = 1.08Solution: Find the opposite side using the pythagorean
theorem (hyp)2 = (adj)2 + (opp)2
(21)2 = (8)2 + (opp)2
=⇒ (opp)2 = 377
=⇒ opp =√
377 = 19. 416
We know,
csc α =hypopp
=21√377
= 1. 081 6
�
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HINT
To find sec α if the side opposite α is 15 unitslong and the hypotenuse is 31 units you willneed the hypotenuse and the side adjacent toα.
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Answer: sec α = 1.14Solution: Find the adjacent side using the pythagorean
theorem (hyp)2 = (adj)2 + (opp)2
(31)2 = (adj)2 + (15)2
=⇒ (adj)2 = 736
=⇒ adj =√
736 = 27. 129
We know,
sec α =hypadj
=31√736
= 1. 142 7
�
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HINT
To find tan α if the side adjacent α is 4.6units long and the hypotenuse is 9.2 unitsyou will need the length of the side oppositeα and the hypotenuse.
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Answer: tan α = 1.73Solution: Find the opposite side using the pythagorean
theorem.
(hyp)2 = (adj)2 + (opp)2
=⇒ (9.2)2 = (4.6)2 + (opp)2
=⇒ (opp)2 = 63. 48
=⇒ opp =√
63.48 = 7. 967 4
We know,
tanα =oppadj
=7.9674
4.6= 1. 732
�
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c
10 9
67α
γ
To find sin α if a = 9, b =10, and β = 67◦ as depictedin the figure, use the law ofsines.
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Answer: sin α = 0.83
c
10 9
67α
γ
Solution: By the law of sinessin α
a=
sin β
b
sin α
9=
sin 67◦
10so
sin α = 9 (0.092 05)= 0.828 45
�
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b
10
14
72
γ
β
To find sin γ in the given trian-gle recall that the law of sines
sin α
a=
sin β
b=
sin γ
c.
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Answer: sin γ = 0.07
b
10
14
72
γ
β
Solution: Use the law of sinessin α
a=
sin γ
c
to obtainsin 72◦
14=
sin γ
10sin γ = 10 (.00679)
= 0.0679�
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To find c if β = 26◦, b = 7, and γ = 75◦ asdepicted in the figure below use the law ofsines.
a7 75
c26α
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Answer: c = 15.42
a7 75
c26α
Solution: By the law of sinessin γ
c=
sin β
b.
Sosin 75◦
c=
sin 26◦
77 sin 75◦ = c sin 26◦
c =7 sin 75◦
sin 26◦ = 15.424�
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To find a in the figure use the law of sines
which equatessin α
ato similar ratio.
a19.4
112c
27
γ
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Answer: α = 9.5
a19.4
112c
27
γ
Solution: By the law of sinessin α
a=
sin β
b.
Sosin 27◦
a=
sin 112◦
19.419.4 sin 27◦ = a sin 112◦
α =19.4 sin 27◦
sin 112◦ = 9. 499�
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HINT
To find cos β for the angle β in the triangle,use the law of cosines. Remember this lawcan be written in three different forms.
1811
9
γ
βα
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Answer: cos β = .88Solution: By the law of cosines (b2 = a2 + c2 − 2ac cos β)
112 = 182 + 92 − 2 (18) (9) cos β
112 = 405 − 324 cos β
=⇒ cos β =121 − 405
−324= 0.876 54
1811
9
γ
βα
�
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To find cos γ if a = 45, b = 42, and c = 70recall the law of cosines:
c2 = a2 + b2 − 2ab cos γ.
4542
70
γ
βα
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Answer: cos γ = −0.29Solution: Using the law of cosines c2 = a2 + b2 − 2ab cos γ
we have
702 = 452 + 422 − 2 (45) (42) cos γ
= 3789 − 3780 cos γ
=⇒ cos γ =4900 − 3789
−3780= −0.293 92
4542
70
γ
βα
�
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To find a if α = 47◦, b = 14, and c = 24 usethe law of cosines which equates
b2 + c2 − 2bc cos α
to another value.
14
2447
aγ
β
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Answer: a = 17.71Solution: By the law of cosines a2 = b2 + c2 − 2bc cos α we
have
a2 = 142 + 242 − 2 (14) (24) cos 47◦
a2 = 772 − 672 cos 47◦
a2 = 313.7
=⇒ a =√
313.7 = 17.712
14
2447
aγ
β
�
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HINT
To find b in the triangle use the form ofthe law of cosines that equates b2 to anothervalue.
b
1.792
2
γ
α
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Answer: b = 2.67
b
1.792
2
γ
α
Solution: By the law of cosines
b2 = a2 + c2 − 2ac cos β
we have
b2 = 22 + (1.7)2 − 2 (2) (1.7) cos 92◦
b2 = 6.89 − 6.8 cos 92◦
b2 = 7.127 3
=⇒ b =√
7.1273 = 2.6697�
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