Trignometry, class 10 Tamil Nadu state board, Unit 6

1. Trigonometry
Unit 6

2. Content
› Trigonometry meaning
› Trigonometric ratios
› Trigonometric ratios of complementary angle
› Trigonometric ratios of Some Specific Angles

3. Trigonometry
› The word ‘trigonometry’ is derived from the Greek words
‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’
(meaning measure). In fact, trigonometry is the study of
relationships between the sides and angles of a triangle.
› Hipparchus is considered as “The Father of
Trigonometry”.

4. Trigonometric ratio
Now we will study some ratios of the sides of a right triangle with
respect to its acute angles, called trigonometric ratios of the angle.
𝑠𝑖𝑛𝑒 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐴
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝐵𝐶
𝐴𝐶
𝑐𝑜𝑠𝑖𝑛𝑒 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠𝐴
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝐴𝐵
𝐴𝐶
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐴
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠ 𝐴
=
𝐵𝐶
𝐴𝐵

5. Trigonometric ratio
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑛𝑒 ∠𝐴
𝑐𝑜𝑠𝑖𝑛𝑒 ∠𝐴
=
𝐵𝐶
𝐴𝐶
𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
𝑐𝑜𝑠𝑖𝑛𝑒 ∠𝐴
𝑠𝑖𝑛𝑒 ∠𝐴
=
𝐴𝐶
𝐵𝐶
Also we see that, 𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
1
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 ∠𝐴
𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
1
𝑠𝑖𝑛𝑒 ∠𝐴
𝑠𝑒𝑐𝑎𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
1
𝑐𝑜𝑠𝑖𝑛𝑒 ∠𝐴

6. Pneumonic to remember ratios
SOH
SINE is OPPOSITE by
Hypotenuse
CAH
COS is ADJACENT by
Hypotenuse
TOA
TANGENT is OPPOSITE by
Adjacent
For Cosec and sec
Remember two “CO” will not
come together.
𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
1
𝑠𝑖𝑛𝑒 ∠𝐴
𝑠𝑒𝑐𝑎𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
1
𝑐𝑜𝑠𝑖𝑛𝑒 ∠𝐴
Remark : Since the hypotenuse is the longest side in a right triangle, the value
of sin A or cos A is always less than 1 (or, in particular, equal to 1).

7. 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒2
= 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡2
+ (𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒2
)
Recall Pythagoras Theorem,

8. PROBLEM 1
𝑠𝑖𝑛𝑒 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐴
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
3
4
Let,
BC= 3X
AC= 4X
Using Pythagoras theorem,
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒2
= 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡2
+ (𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒2
)
(4X2)= (adjacent2) + (3X2)
16X2 = (adjacent2) + 9X2
16X2 – 9X2 = (adjacent2)
7X2 = (adjacent2)
Adjacent = 7𝑋
If sin A = 3 /4 calculate
cos A and tan A.

9. PROBLEM 1
Now we find cos A and Tan A
𝑐𝑜𝑠𝑖𝑛𝑒 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠𝐴
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
𝐴𝐵
𝐴𝐶
=
7𝑋
4𝑋
=
7
4
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 ∠ 𝐴 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐴
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠ 𝐴
=
𝐵𝐶
𝐴𝐵
=
3𝑋
7𝑋
=
3
7
If sin A = 3 /4 calculate
cos A and tan A.

10. In Δ ABC, right-angled at B, AB = 24 cm, BC =
7 cm. Determine sin A, cos A
Answer:
𝑆𝑖𝑛 𝐴 =
7
25
𝐶𝑜𝑠 𝐴 =
24
25

11. Trigonometric ratios of complementary angle
› The sine of any acute angle is equal to the cosine of its
complement and vice versa.
› The cotangent of any acute angle is equal to the tangent
of its complement and vice versa.
› The cosecant of any acute angle is equal to the secant of
its complement and vice versa.
If the sum of two angles is one right angle or 90°, then
one angle is said to be complementary of the other.

12. Derivation
Consider a semicircle of radius 1 as shown in the
figure.
Let QOP = 𝜃.
Then QOR =90°− 𝜃, so that OPQR forms a
rectangle.
From triangle OPQ , OP / OQ = cos 𝜃
But OQ = radius = 1
∴ OP =OQ cos 𝜃 = cos 𝜃
Similarly, PQ/ OQ = sin 𝜃
PQ =OQ sin 𝜃 = sin 𝜃 ( OQ = 1)
OP = cos 𝜃, PQ = sin 𝜃 … (1)

13. Derivation
Now, from triangle QOR,
we have OR/OQ = cos(90°−𝜃)
∴ OR =OQ cos(90°−𝜃)
OR =cos(90°−𝜃)
Similarly, RQ/OQ = sin(90°−𝜃)
Then, RQ = sin(90°−𝜃)
OR= cos(90°− 𝜃), RQ = sin(90°−𝜃) … (2)
OPQR is a rectangle,
OP = RQ and OR = PQ
Therefore, from (1) and (2) we get,
sin(90°− 𝜃) = cos 𝜃 and cos(90°− 𝜃) = sin 𝜃

14. Trigonometric Ratios of Some Specific Angles
Trigonometric Ratios of 30° and 60°
Let us now calculate the trigonometric ratios of 30°
and 60°. Consider an equilateral triangle ABC. Since
each angle in an equilateral triangle is 60°, therefore,
∠ A = ∠ B = ∠ C = 60°.
Draw the perpendicular AD from A to the side BC.
Now Δ ABD ≅ Δ ACD
Therefore, BD = DC
and ∠ BAD = ∠ CAD (CPCT)
Now observe that:
Δ ABD is a right triangle, right-angled at D with
∠ BAD = 30° and ∠ ABD = 60°
3
6

15. Trigonometric Ratios of Some Specific Angles
Trigonometric Ratios of 30° and 60°
Using Pythagoras theorem,
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒2
= 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡2
+ (𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒2
)
22
= 12
+ (ℎ2
)
ℎ = 3
𝑆𝑖𝑛 30° =
1
2
𝑆𝑖𝑛 60° =
3
2
𝐶𝑜𝑠 30° =
3
2
𝐶𝑜𝑠 60° =
1
2
tan 30° =
1
3
tan 60° =
3
1
3
6

16. Trigonometric Ratios of Some Specific Angles
› To find out trigonometric ratios of 45° we consider an
isosceles right triangle and proceed.

17. TRIGONOMETRIC
RATIOS OF SOME
SPECIFIC ANGLES
TIPS:
1. Cos 𝜃 will be reverse
order of Sin 𝜃.
2. tan 𝜃 will be sin 𝜃 divided
by cos 𝜃.
3. Cosec 𝜃, sec 𝜃 and cot
𝜃 will be reciprocal of sin
𝜃, cos 𝜃 and tan
𝜃 respectively.
REMEMBER only for Sin.