Introduction to Trigonometry Practice more on Introductions to Trigonometry Page - 1 www embibe com CBSE NCERT Solutions for Class 10 Mathematics
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Class- X-CBSE-Mathematics Introduction to Trigonometry
Practice more on
Introductions to
Trigonometry
Page - 1 www.embibe.com CBSE NCERT So
lutions f or C lass 10Mathematics
Chapter
Back of Chapter Questions
1.In ο
ABC , right-angled at B,AB=24 cm,BC= 7 cm. Determine: (i)sinA,cosA (ii)sinC,cosCSolution:
In ȟABC, apply Pythagoras theorem
AC = AB + BC = (24) + (7) =625AC=ξ625
=25 cm (i)sinA = =7 25cosA
Side adjacent to ס
hypotenuse =AB AC =24 25(ii) Class- X-CBSE-Mathematics Introduction to Trigonometry
Practice more on
Introductions to
Trigonometry
Page - 2 www.embibe.com
sinC =Side opposite to ס hypotenuse=AB AC=24 25cosC =Side adjacent to ס hypotenuse =BCAC=7 25
2. In Fig., find tanPെcotR .
Solution:
Apply Pythagoras theorem in ȟPQR
PR = PQ + QR 13) =(12) +QR169=144+QR
25= QR
QR= 5 tanP =Side opposite to סSide adjacent to ס
=QR PQ 5 12 cotR =Side adjacent to סSide opposite to ס
=QR PQ 5 12 tanPെcotR =5 12 െ5 12= 0 Class- X-CBSE-Mathematics Introduction to TrigonometryPractice more on
Introductions to
Trigonometry
Page - 3 www.embibe.com
3. If sinA =
, calculate cosA and tanA.Solution:
Let us assume that ȟ ABC is a right triangle, right angled at vertex B.Given that
sinA =3 4 sinA = BC AC =34 Let BC and AC be 3R and 4R respectively, where R is any positive number.In ȟABC, apply Pythagoras theorem
AC = AB + BC (4 R) = AB +(3 R) െ9 R = AB = AB R cosA =Side adjacent to ס hypotenuse AB AC =ξ7 R 4R =ξ7 4 tanA =Side opposite to סSide adjacent to ס
BC AB =3R 7 R =3 74. Given 15cotA =8 , find sinA and secA.
Solution:
Let us assume that ȟABC is a right triangle, right angled at vertex B. Class- X-CBSE-Mathematics Introduction to TrigonometryPractice more on
Introductions to
Trigonometry
Page - 4 www.embibe.com
cot A= Side adjacent to סSide opposite to ס
AB BC cotA = (given) AB BC =8 15 Let AB and BC be 8R and 15R respectively, where R is a positive number.Now applying Pythagoras theorem in
ȟABC
AC = AB + BC (8R) +(15R) =64 R +225 R=289 R sinA =Side opposite to ס hypotenuse =BC AC 15 R 17 R =1517 secA =hypotenuse
Side adjacent to ס
AC AB =17 85. Given secɅ=
, calculate all other trigonometric ratios.Solution:
Let us assume that ȟABC is a right angled triangle, right angled at vertex B. Class- X-CBSE-Mathematics Introduction to TrigonometryPractice more on
Introductions to
Trigonometry
Page - 5 www.embibe.com
secɅ=hypotenuseSide adjacent to ס
13 12 =AC AB Let AC and AB be 13R and 12R, where R is a positive number.Now applying Pythagoras theorem in
ȟABC
AC) =(AB) +(BC) (13 R) =(12 R) +BC =144 R + BC = BC sinɅ=Side opposite to ס hypotenuse =BCAC=5R 13R=5 13 cosɅ=Side adjacent to ס hypotenuse =ABAC=12R13R=1213
tanɅ=Side opposite to סSide adjacent to ס
=BC AB=5R 12R=5 12 cotɅ=Side adjacent to סSide opposite to ס
=ABBC=12R
5R=12 5 cosesɅ=hypotenuseSide opposite to ס
=ACBC=13R 5R=13 56. If סA and סB are acute angles such that cosA =cosB, then show that סA =ס
Solution:
Let us assume a right triangle, right angled at
vertex C. Class- X-CBSE-Mathematics Introduction to TrigonometryPractice more on
Introductions to
Trigonometry
Page - 6 www.embibe.com
Since cosA =cosB
SoSo סA =ס
7. If cotɅ=
, evaluate: (i) (ii) cotSolution:
Let us assume that ȟ ABC is a right triangle, right angled at vertex B. cotɅ= Side adjacent to סSide opposite to ס
=BC AB 7 8 Let BC and AB be 7R and 8R respectively, where R is a positive number.In ȟABC, apply Pythagoras theorem
AC = AB + BC (8 R) +(7 R) =64 R +49 R=113 R
AC=ξ113
R Class- X-CBSE-Mathematics Introduction to TrigonometryPractice more on
Introductions to
Trigonometry
Page - 7 www.embibe.com
sinɅ=Side opposite to ס hypotenuse =AB AC 8 R 113R =8 113
cosɅ=Side adjacen to ס hypotenuse =BC AC cosɅ= 7 R 113
R =7 113
(i) =1െ ൬8 113
1െ ൬7
113൰
=1െ64 1131െ49
113113െ64
113113െ49
113=4964 (ii) cot =(cotɅ) A 6