4 Material AICLE. 4º de ESO: Trigonometry (Solucionario)
1.
2.
3.
trigonometricangle measurechordsine functiontriangulationtrigonometric series
5Material AICLE. 4º de ESO: Trigonometry (Solucionario)
4.
5.
SINECOSINE
TANGENTCOSECANT
SECANTCOTANGENT
Opposite / AdjacentOpposite / HypotenuseHypotenuse / OppositeAdjacent / Hypotenuse
Adjacent / OppositeHypotenuse / Adjacent
SINE RULECc
Bb
Aa
sinsinsin==
side a divided by the sine of angle A equals side b divided by the sine of angle B equals side c divided by the sine of angle C
COSINE RULE Abccba cos2222 −+=
side a squared equals side b squared plus side c squared minus twice b times c times the cosine of angle A
6 Material AICLE. 4º de ESO: Trigonometry (Solucionario)
7.A. For the figure given on the left, the value of sin C is
Answer c / bThe sine of an angle is defined as Opposite Side / Hypotenuse. Now for angle C, the oppositeside is c and the hypotenuse is b. Hence the correct answer is c/b.
B. From the figure given on the right, the value of sin A + cos A is
Answer (a + c)/bWe know sin A = a/b, and cos A = c/b. Hence sin A + cos A = (a + c)/b.
C. From the figure given on the left, the value of cos C is
Answer a / bWe know that cos of any angle = Base/Hypotenuse. Now for the angle C, the base is a andhypotenuse is b. So cos C = a/b.
D. For the figure given on the right, which of the following relationships is true :
Answer cot A = c / aBy definition, cot A = 1 / tan A = c / a.
E. From the figure given on the left, the value of cos C + sin A is
Answer 2a/bThe value of cos C = a/b. Similarly the value of sin A = a/b. Hence cos C + sin A = 2a/b.
F. Which of the following relationships is true:
Answer sin A / cos A = tan AThe expression sin A / cos A = tan A is a useful one to remember in trigonometry.
G. tan A / sin A =
Answer sec Atan A = sin A / cos A. Therefore tan A / sin A = 1 / cos A = sec A.
H. (sin A / tan A) + cos A =
Answer 2 cos AWe know tan A = sin A / cos A. Therefore (sin A / tan A) + cos A = cos A + cos A = 2 cos A.
I. cot A tan A =
Answer 1cot A = 1 / tan A. Hence cot A tan A = 1.Alternatively cot A = cos A/sin A and tan A= sin A/cos A. So cot A tan A = (cos A/sin A) (sin A/cos A) = 1.
J. From the figure, the value of cosec A + cot A is:
Answer (b + c)/aWe know cosec A = b/a and cot A = c/a. Hence cosec A + cot A = (b + c)/a.
7Material AICLE. 4º de ESO: Trigonometry (Solucionario)
8.
Example 1 Example 2 Example 3
K. Which of the following relationships is true:
Answer cos A sec A = 1By definition, sec A = 1 / cos A. So cos A sec A = 1 is true.
L. From the figure, the value of sin2 A + cos2 A is
Answer 1This question is a bit tricky. We know sin A = a/b and cos A = c/b. So sin2 A + cos2 A = (a2 + c2)/ b2. By Pythagoras Theorem, a2 + c2 = b2 for a right-angled triangle. Hence sin2 A + cos2 A = 1,which is a famous identity.
M. From the figure, the value of cot C + cosec C is
Answer (a + b)/ccot C is Base/Opposite Side and cosec C is Hypotenuse/Opposite Side. From thesedefinitions, the values of cot C and cosec C are given by a/c and b/c respectively. Hencethe answer is (a + b)/c.
N. cosec A / sec A =
Answer cot ABy definition, cosec A = 1 / sin A and sec A = 1 / cos A. So cosec A / sec A = cos A / sin A =cot A.
O. For the figure given on the right, the value of cot A is
Answer tan CThe value of cot A is c/a. Similarly the value of tan C is c/a. Hence cot A = tan C.
8 Material AICLE. 4º de ESO: Trigonometry (Solucionario)
9.
Example 4 Example 5
A.
(a) a = 2, A = 30°, B = 40°b = 2.571, c = 3.75
(b) b = 5, B = 45°, C = 60°a = 7.044, c = 6.124
(c) c = 3, A = 37°, B = 54°a = 1.806, b = 2.427
B.
(a) a = 3, b = 5, A = 32°Two possible triangles:B = 62°, C = 86°, c = 5.647and B = 118°, C = 30°, c = 2.833
(b) b = 2, c = 4, C = 63°B = 27°, A = 88°, a = 4.487
(c) c = 2, a = 1, B = 108°b = 2.457, A = 23.2°, C = 51.9°
C.
(a) a = 1, b = 2, C = 30°c = 1.239, A = 23.8°, B = 126.2°
(b) a = 3, c = 4, B = 50°b = 3.094, A = 48°, C = 82°
(c) b = 5, c = 10, A = 30°a = 6.197, B = 23.8°, C = 126.2°
D.
(a) a = 2, b = 3, c = 4A = 29.0°, B = 46.6°, C = 104.5°
(b) a = 1, b = 1, c = 1.5A = 41.4°, B = 41.4°, C = 97.2°
9Material AICLE. 4º de ESO: Trigonometry (Solucionario)
11.
15.
16.
a) The distance of the man from the tower is 20.21 mb) The length of the string used by the little boy is l = 2 h = 2 (15) = 30 mc) The height of the second tower is 46.19 md) The distance of the ship from the lighthouse is 35.49 me) The velocity of the plane is given by V = distance covered / time takenV= DE / 60 = 19.25 m/s
tan A = 1.23A = 50.9 °
tan B = 2.56B = 68.7 °
sin C = 0.78C = 51.3 °
sin D = 0.527D = 31.8 °
cos E = 0.352E = 69.4 °
cos F = 0.725F = 43.5 °
tan G = 0.786G = 38.2 °tan H = 1.275H = 51.9 °
sin I = 0.468I = 27.9 °
sin J = 0.867J = 60.1 °
sinA cosA tanA secA cosecA cotA0º 0 1 0 1 none none30º 0.5 0.8660 0.5773 1.1547 2 1.732045º 0.7071 0.7071 1 1.4142 1,4142 160º 0.8660 0.5 1.7320 2 1.1547 0.577390º 1 0 none none 1 0
10 Material AICLE. 4º de ESO: Trigonometry (Solucionario)
17.
Score
Requirement 0 1 2 3 4
Story No Story or logical sequence of thoughts. One sentence.
No real story.Lacks imaginationand thought. No
real application ofa trigonometry
problem. Spellingand grammar
mistakes.
Story lacks one ofthe following:imagination,
completesentences,
complete thought,but still involves a
trigonometryproblem. Somegrammar and
spelling mistakes.
Imaginative storycomprised of
mostly completesentences that
involves atrigonometry
problem. Somegrammar mistakes
or a fewmisspellings.
Imaginative storycomprised of
completesentences that
involves atrigonometryproblem. No
grammar mistakesor misspellings.
Picture ordrawing
Nothing clearlydefinable, or
understandable
Picture lacks atleast two of the
following: illustratesthe story, clarity,appropriate size,color, appropriate
subject matter.The picture is not
very visuallyappealing.
Picture lacks atleast one of the
following: illustratesthe story, clarity,appropriate size,and appropriatesubject matter.
The picture is stillvisually appealing
and has color.
Creative picture ordrawing thatsomewhat
illustrates the storyand the
trigonometryproblem. Mainly
clear, colorful andvisually appealing.Appropriate size
and topic.
Creative picture ordrawing that
clearly illustratesthe story and the
trigonometryproblem. Clear,
colorful andvisually appealing.Appropriate size
and topic.
Diagram oftriangle
No serious attempt to make diagram.
Lacks labels, units and accuracy to story.
Not drawn with astraight edge orcomputer, and
missing ONE of thefollowing: units,labels. Correctlydrawn to match
story, picture andsolution.
Clearly drawnusing a straightedge
or computer.Missing TWO of the
following: units,labels. Correctlydrawn to match
story, picture andsolution.
Clearly drawnusing a straightedge
or computer.Missing ONE of the
following: units,labels. Correctlydrawn to match
story, picture andsolution.
Clearly drawnusing a straightedge
or computer.Units included,
labeled with rightangle. Correctlydrawn to match
story, picture andsolution.
Calculations
Not neatly typed or written and
missing TWO of thefollowing: units, formulas,
eachstep of the problem, correct
answer.
Not neatly typed orwritten and missing
ONE of thefollowing: units,formulas, each
step of theproblem, correct
answer.
Neatly typed orwritten. Missing
TWO of thefollowing: units,formulas, each
step of theproblem, correct
answer.
Neatly typed orwritten. Missing
One of thefollowing: units,formulas, each
step of theproblem, correct
answer
Neatly typed orwritten. Formulaslisted. Each stepclearly outlinedand included.
Units included inanswer. Correct
answer to problem
Presentation Put together on lined paper or notebook paper.
Rushed,incomplete, notprofessional.
Not quiteprofessional. Looksrushed or quickly
put together.Lacks colors, title or
does not fitstandard sizerequirements
Neat andProfessional buthas ONE of the
following: Does notuse 3 colors, title
not included, not astandard size.
Professional look(no tape showing,neatly constructed
or drawn)Story should betyped or neatly
printedAt least 3 Colors Title should be
obvious and neatSize limit: 8½ x 11
inches or a standard sheet of
construction paper.