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Published on: 24/09/2019
Term 3 Trigonometry
Download Tamil Nadu 9th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Find the value of \(\cfrac { cos{ 63 }^{ 0 }20' }{ sin{ 26 }^{ 0 }40' } \)
2.
Find the value of \(\frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}}\)
3.
Find the value of cot 15°. cot 30°. cot 45°. cot 60°. cot 75°
4.
If 3 (tan \(\theta\)) + 4 (sec \(\theta\) \(\times\) sin 6) = 24. Then find all the trigonometric ratios of the angle \(\theta\)
5.
If sin\(\theta\) = \(\cfrac { a }{ \sqrt { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc \right) } } \), then show that (b + c) sin \(\theta\) = (a) cos \(\theta\)
6.
If 3 cot\(\theta\) = 1, then find the value of \(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } \)
7.
Find the angle made by a ladder of length 5m with the ground, if one of its end is 4 m away from the wall and the other end is on the wall.
8.
Find the area of a right triangle whose hypotenuse is 10cm and one of the acute angle is 24024'
9.
Find the value of \(\theta\) if
(i) sin \(\theta\) = 0.9975
(ii) cos \(\theta\) = 0.6763
(iii) tan \(\theta\) = 0.0720
(iv) cos \(\theta\) = 0.0410
(v) tan \(\theta\) = 7.5958
10.
Find the value of 8 sin 2x cos 4x sin 6x , when x =150
11.
Verify 3 cos A = 4 cos3 A - 3 cosA , when A = 300
12.
If sin \(\theta\) = \(\frac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \) then show that b sin \(\theta\) = a cos \(\theta\)
13.
If cos A = \(\frac { 2x }{ 1+{ x }^{ 2 } } \) then find the values of sinA and tan A in terms of x.
14.
From the given figure, find the values of
(i) sin B
(ii) sec B
(iii) cot B
(iv) cos C
(v) tan C
(vi) cosec C
15.
From the given figure, find all the trigonometric ratios of angle B.
1.
\(\cfrac { cos{ 63 }^{ 0 }20' }{ sin{ 26 }^{ 0 }40' } =\cfrac { cos63^{ 0 }20' }{ cos{ 63 }^{ 0 }20' } =1\)
2.
\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)
3.
cot (90° - 75) cot (90° - 60°) cot 45° cot 60° cot 75°
= tan 75° tan 6.0° (1) cot 60° cot 75° = 1
4.
3 tan \(\theta\) + 4 (sec \(\theta\) \(\times\) sin\(\theta\)) = 24
\(3tan\theta +4\left( \cfrac { 1 }{ cos\theta } \times sin\theta \right) =24\)
7 tan \(\theta\) = 24
\(tan\theta =\cfrac { 24 }{ 7 } \)
hypetenuse = \(\sqrt { { 24 }^{ 2 }+49 } =\sqrt { 576+49 } =\sqrt { 625 } =25\)
\(sin\theta =\cfrac { 24 }{ 25 } ;cos\theta =\cfrac { 7 }{ 25 } ;cosec\theta =\cfrac { 25 }{ 24 } ;sec\theta =\cfrac { 25 }{ 27 } ;cot\theta =\cfrac { 7 }{ 24 } \)
5.
,Adjacent side = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc-{ a }^{ 2 } } \)
= \(\sqrt { \left( b+c \right) ^{ 2 } } =b+c\)
\(\therefore\) \((b+c)sin\theta =\alpha cos\theta \)
\(\left( b+c \right) \times \cfrac { a }{ \sqrt { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc \right) } } =a\times \cfrac { b+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc } } \)
6.
\(cot\theta =1\)
\(cot\theta =\cfrac { 1 }{ 3 } \)
\(\cfrac { adjacent }{ opposite } =\cfrac { 1 }{ 3 } \)
\(\sqrt { { 3 }^{ 2 }+1 } =\sqrt { 10 } \)
\(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } =\cfrac { 3\times \cfrac { 1 }{ \sqrt { 10 } } -4\times \cfrac { 3 }{ \sqrt { 10 } } }{ 5\times \cfrac { 3 }{ \sqrt { 10 } } +4\times \cfrac { 1 }{ \sqrt { 10 } } } =\cfrac { 3-12 }{ 15+4 } =\cfrac { -9 }{ 19 } \)
7.
\(cos\theta =\cfrac { 4 }{ 5 } =0.8\)
cos 36° 48' = 0.8
\(\therefore\) \(\theta\) = 36048'
8.
Hypotenuse = 10 cm
One of the acute angle = 24° 24'
sin 24° 24' = 0.4131'
\(\cfrac { x }{ 10 } =0.4131\)
x = 0.4131 \(\times\) 10
x = 4.131
cos 24° 24' = 0.9107
\(\cfrac { y }{ 10 } =0.9107\)
y = 9.107
\(\therefore\) Area of the triangle = \(\cfrac { 1 }{ 2 } bh\)
= \(\cfrac { 1 }{ 2 } \times y\times x\)
\(=\cfrac { 1 }{ 2 } \times 9.107\times 4.131=18.81sq.cm\)
9.
(i) From the natural sines table
sin 85° 57' = 0.9975
\(\therefore\) \(\theta \) = 85° 57'
(ii) cos\(\theta \) = 0.6763
cos 47° 33' = 0.6762
\(\therefore\) \(\theta \) = 47° 33'
(iii) tan \(\theta \) = 0.0720
tan 4° 7' = 0.0720
\(\therefore\)\(\theta \) = 4°7'
(iv) cos \(\theta \) = 0.0410
cos 87° 45° = 0.0410
:\(\therefore\)\(\theta \) = 87°39'
(v) tan \(\theta \) = 7.5958
tan 82° 30' = 7.5958
\(\therefore\) \(\theta \) = 82° 30'
10.
8 sin 2(15°). cos 4 (15°). sin 6(15°)
= 8 sin 30° cos 60° sin 90°
= \(8\times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times 1=2\)
11.
L.H.S = cos 3A= cos 3 (30°)
= cos 90°
=0 ---- (1)
R.H.S = 4 cos3 A- 3 cosA
= 4 cos3 30° - 3 cos 30°
= \(4\left( \cfrac { \sqrt { 3 } }{ 2 } \right) ^{ 3 }-3\times \cfrac { \sqrt { 3 } }{ 2 } \)
(1) = (2). Hence it is verified
12.
\(sin\ \theta =\cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(cos\ \theta =\cfrac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(b\ sin\ \theta =b\times \cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } -\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \) ... (1)
\(a\ cos\ \theta =a\times \cfrac { b }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } } =\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \) ...(2)
(1) = (2)\(\Rightarrow\) LHS = RHS
Hence proved
By By the Pythagoras theorem
BC2 = AC2 + AB2
= \(\left( \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \right) ^{ 2 }-{ a }^{ 2 }\)
= a2+b2-a2
= b2
BC = b
13.
By the pythagoras theorem,
AB2 = OA2 + OB2
(1 + x2)2 = (2x)2 + OB2
OB2 = (1 +x2)2 - (2x)2 = 1+ x4 + 2x2 - 4x2 = 1+x4 - 2x2
OB2 = (1-x2)2 B
OB = (1-x2)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)
14.
(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
= \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv)
(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi) \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)
By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\)
15.
sin B = \(\frac { 9 }{ 41 } \);
cos B = \(\frac { 40 }{ 41 } \);
tan B =\(\frac { 9 }{ 40 } \) ;
cosec B = \(\frac { 1 }{ sinB } \) = \(\frac { 41 }{ 9 } \);
sec B =\(\frac { 1 }{ cotB } \) = \(\frac { 41 }{ 40 } \);
cot B =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)
9th Standard Syllabus & Materials
9th Standard
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
