T-ratios of Standard Angles
|
0° |
|
|
|
|
sin
|
0 |
1/2 |
1/2 |
(3)/2 |
1 |
cos
|
1 |
(3)/2 |
1/2 |
1/2 |
0 |
tan
|
0 |
1/3 |
1 |
3 |
not defined |
cot
|
not defined |
3 |
1 |
1/3 |
0 |
sec
|
1 |
2/3 |
2 |
2 |
not defined |
cosec
|
not defined |
2 |
2 |
2/3 |
1 |
For memorising, we can use the following table:
Illustrative Examples
Example
Solve for a and c in the given triangle. Also find the area
of the ABC.
Solution
sin A = height / hypotenuse = BC/AC
=> sin 30° = a/12
=> a = 12 sin 30° = 12.(1/2)= 6
Similarly cos A = AB/AC => cos 30° = c/12
=> c = 12 cos 30° = 12.(3)/2 = 63
Area of ABC = (1/2) x base x height = (1/2) x c x a = (1/2)
x 63 x 6
= 183 sq. units.
Example
If A, B, A +B, A -B are positive acute angles, find the
values of A and B from the equations:
sin (A -B) = 1/2, cos (A +B) = 1/2
Solution
The given equations are
sin (A -B) = 1/2 = sin 30° => A -B = 30°
...(i)
cos (A +B) = 1/2 = cos 60° => A +B = 60°
...(ii)
Solving (i) and (ii) simultaneously, we get
A = 45°, B = 15°
Exercise
- Show that
(i) sin 30° cos 0° +sin 45° cos 45° +sin 60° cos 30° = 7/4
(ii) 4 sin (/6) sin²(/3) +3 cos(/3)
tan (/4) +cosec²(/2) = 2 sec² (/4)
- Evaluate sec 30° tan 60° +sin 45° cosec 45° +cos 30° cot 60°
- Taking A = 30°, verify that
(i) sin2 A +cos 2 A = 1 (ii) sin 3 A = 3 sin A -4 sin³ A
(iii) sin 2 A = (2 tan A)/(1 -tan² A)
(iv) cos 2 A = (1 -tan² A)/(1 +tan²A)
- Taking A = 60°, B = 30°, verify that
(i) sin (A +B) sin A +sin B
(ii) cos (A +B) cos A +cos B
- Find the value of (0° < <
90°) satisfying
(i) cos / (cosec +1) + cos
/ (cosec -1) = 2
(ii) (cos² -3cos +2)/ sin²
= 1
(iii) 2 sin² = 1/2
(iv) 3 cos = 2 sin²
(v) 2 cosec = 3 sec²
(vi) tan + cot = 2
(vii) sec² = 1 + tan
- Assuming A, B, A +B, A -B to be positive acute angles, find A and B when
(i) sin (A +B) = (3)/2, cos (A -B) = (3)/2
(ii) tan (A +B) = 3, tan (A -B) = 1
Answers
2. 7/2
5. (i) 45° (ii) 60°
(iii) 30° (iv) 60°
(v) 30° (vi) 45°
(vii) 45° (discarding = 0° as 0° <
< 90°)
6. (i) A = 45°, B = 15° (ii) A = 52·5°, B = 7·5°