
Class 10 Trignometry
Presentation
•
Mathematics
•
10th Grade
•
Hard
Standards-aligned
Qaulity Education
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21 Slides • 8 Questions
1
by Qaulity Education
2
Basic trignometry
Trignometry ratios of common angles
Trignometry ratios of complement angle
Trignometry identities
Question
3
Trignometry can only apply to right angle triangle.
Height is side taken opposite to angle taken.
Base is side on which angle taken lie,
Hypotenuse is the longest side of a right-angled triangle.
Sin,cos ratios can never be an improper fraction i.e. denominator is always larger than numerator.
Sec,cosec can never be a proper fraction.
Learn the ratios in order Sin,Cos,Tan.
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5
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1st ratio is of sin that is L/K or P/H remember the order of ratios and trick
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8
9
Multiple Choice
Find trignometry ratios sin x,cos x,tan x in following figure
654 , 657 , 47
657 , 654 , 74
654 , 657 , 74
None of these
10
Multiple Choice
find tan P – cot R.
0
-119/60
119/60
60/119
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12
Hence proved.
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Prove that cot x = cos x/sin x
Ex 8.1 NCERT.
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Learn the angles in order 0,30,45,60,90.
Now as per order take the first line as sin.
Now write 0,1,2,3,4 in order beneath angles.
Now divide the raw in 4 i.e. 0, 1/4 , 2/4 , 3/4 , 4/4
Than take square root of whole raw i.e. 0 , 1/2 ,
6. This is the first raw that is sin function at diffrent angles.
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7. Now write inverse of first raw to find cos.
8. Now divide first and second raw for tan.
9. And the other threes are reciprocal of first three so basically reciprocal
Q.1 Find sin 45 + cos 45.
Q2.Find tan 60-cot 30.
Do NCERT ex 8.2
17
Multiple Choice
cosec260−sec2 0sin30cos230−tan260sin30 =
16
0
-9
9
18
Multiple Choice
tan 45 − tan 60cos30−sin 30 =
23
23
32
0
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Similiarly find all other
ratios
Do ex. 8.3 NCERT
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Multiple Choice
tan 48° tan 23° tan 42° tan 67° =
0
UNDEFINED
NONE
1
22
Multiple Choice
tan 2A = cot (A – 18°),find A
-18
36
18
30/60
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Prove
Similiarly divide eq. with
for II eq. and for III
eq. to prove it.
25
Multiple Choice
9sec2A−9tan2A
cos A
9
1
0
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Multiple Choice
(1+tan x+sec x)(1+cot x - cosec x)
2
2 cos x sin x
0
1
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