4th

1. Triangle
Trigonometry

2. SOH CAH TOA

4. Length of Sides (in cm) Trigonometric Ratios
Angle
(°)
Opposite Adjacent Hypotenuse 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
a.
b.
c.
d.
e.
f.

5. Length of Sides (in cm) Trigonometric Ratios
Angle
(°)
Opposite Adjacent Hypotenuse 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
a. 14° 2cm 7.8cm 8.2cm 0.24 0.97 0.25
b. 24° 3.3cm 5.9cm 7.4cm 0.41 0.91 0.45
c. 54° 6.2cm 5.5cm 7.3cm 0.81 0.59 1.38
d. 65° 5.1cm 2.4cm 5.7cm 0.91 0.42 2.14
e. 30° 3.5cm 4.9cm 6.7cm 0.5 0.87 0.58
f.51° 5.6cm 4.7cm 7.2cm 0.78 0.63 1.23

6. 𝑺𝒊𝒏𝜽 =
𝒐𝒑𝒑(𝒃)
𝒉𝒚𝒑(𝒄)

7. 𝒄𝒐𝒔𝜽 =
𝒂𝒅𝒋(𝒂)
𝒉𝒚𝒑(𝒄)

8. 𝒕𝒂𝒏𝜽 =
𝒐𝒑𝒑(𝒃)
𝒂𝒅𝒋(𝒂)

9. Example 1: Triangle BCA is a right-
angle at C. If 𝑐 = 23 and 𝑏 =
17, find angle A, angle B and a.
Express your answers up to two
decimal places.

10. cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑏
𝑐
cos 𝐴 =
17
23
𝑐𝑜𝑠 𝐴 = 0.7391
𝐴 = cos−1
(0.7391)
𝐴 = 42.34°

11. b. Since in part (a), it was already
found that
∠𝐴 = 42.34°,
Then
∠𝐵 = 90° − 42.34°
∠𝐵 = 47.66°

12. c. Using the Pythagorean Theorem:
𝑎2
+ 𝑏2
= 𝑐2
𝑎2
+ 17 2
= 23 2
𝑎2
+ 289 = 529
𝑎2
= 529 – 289
𝑎2
= 240
𝑎 = 240
𝑎 = 15.49

13. Example 2: Triangle BCA is
right-angle at C if 𝑐 = 27 and
∠𝐴 = 58°, find ∠𝐵, b and a.
Solution:
a. To find B, since ∠𝐵 𝑎𝑛𝑑 ∠𝐴 are
complementary angles, then

14. 58°
∠𝐵 + ∠𝐴 = 90°
∠𝐵 = 90° − 58°
∠ 𝐵 = 32°

15. b. To find b, since b is the
adjacent side of ∠A and c is the
hypotenuse of right ΔBCA then
use CAH.

16. cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑏
𝑐
cos 58° =
𝑏
27
𝑏 = 27 cos 58°
𝑏 = 27(0.5299) =14.31

17. c. To find a, since a is the opposite side of
∠A then use________?
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑠𝑖𝑛 𝐴 =
𝑎
𝑐
sin 58° =
𝑎
27
𝑎 = 27 𝑠𝑖𝑛 58°
𝑎 = 27 0.8480
𝑎 = 22.90
SOH.

18. Ready?

19. 1. Triangle ACB is right-angled at
C. if ∠𝐴 = 63° and 𝑎 = 11 𝑐𝑚,
find ∠B, b and c.
2. Triangle ACB is right-angled at C. If
𝑎 = 18.5 𝑐𝑚 and 𝑏 = 14.2 𝑐𝑚,
find c, ∠A and ∠B.

20. Solution in
#1

21. a. To find B, take
note that B and A
are complementary
angles, then
∠𝐵 + ∠𝐴 = 90°
∠𝐵 = 90 °– 63°
∠𝐵 = 27°

22. b. To find b, since b is the adjacent side and a is the
opposite side of ∠𝐴, then use TOA.
𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
tan 𝐴 =
𝑎
𝑏
tan 63° =
11
𝑏
𝑏 tan 63° = 11
𝑏 1.9626 = 11
𝑏 =
11
1.9626
𝑏 = 5.60𝑐𝑚
b
27°

23. c. To find c, since c is the hypotenuse and a is
the is opposite side of ∠𝐴, then use SOH.
𝑠𝑖𝑛 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑠𝑖𝑛 𝐴 =
𝑎
𝑐
sin 63° =
𝑎
27
𝑎 = 27 sin 63°
𝑎 = 27 (0.8910)
𝑎 = 24.0571 or 24.06

24. Solution in
#2

25. a. To find c, use Pythagorean
theorem
𝑐2
= 𝑎2
+ 𝑏2
𝑐2
= 18.5 2
+ 14.2 2
𝑐2
= 342.25 + 201.64
𝑐2
= 543.89
𝑐 = 543.89
𝑐 = 23.32

26. b. To find ∠A, since a and b are opposite and
adjacent side of ∠A respectively, then use TOA.
𝑡𝑎𝑛 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑡𝑎𝑛 𝐴 =
𝑎
𝑏
𝑡𝑎𝑛 𝐴 =
18.5
14.2
𝑡𝑎𝑛 𝐴 = 1.3028
𝐴 = tan−1
(1.3028)
𝐴 = 52°

27. c. Based on the fact that ∠A
and ∠B are complementary,
the measure of angle ∠B is
90° – 52° = 38°.

28. This lesson was about six trigonometric ratios.
Various activities were provided to help students
illustrate and define the six trigonometric ratios.
They have also learned how to use these concepts in
finding the missing sides and angles of a right
triangle and applied them to real life situations.
Their knowledge in this lesson will help them
understand the next topic, which is, the
trigonometric ratios of special angles.

29. (Things to Remember)
The sin and 𝜽 are married to each
other. They are not treated as
product of sin and 𝜽. They are
inseparable.𝒔𝒊𝒏 𝜽 ≠ 𝒔𝒊𝒏 × 𝜽 This
is TRUE for all trigonometric ratios.

30. Exercise #4.__Directions: Determine
the other side, given that two sides
are:
1. 𝑎 = 5𝑓𝑡., 𝑏 = 10𝑓𝑡., 𝑐 =?
2. 𝑎 = 4𝑘𝑚, 𝑏 =? 𝑐 = 12𝑘𝑚
3. 𝑎 =?, 𝑏 = 8𝑚, 𝑐 = 24𝑚
4. 𝑎 =?, 𝑏 = 6𝑖𝑛𝑐ℎ, 𝑐 = 9𝑖𝑛𝑐ℎ

32. 1.c = 11.18𝑓𝑡
2.b= 11.31𝑘𝑚
3.a= 22.63𝑚
4.a= 6.71𝑖𝑛𝑐ℎ

33. Exercise #4.__
Using the Calculator to find
Trigonometric Ratios
Find the value of the following,
correct to two decimal places.
a. 𝑐𝑜𝑠 23° b.𝑠𝑖𝑛 65° c.𝑡𝑎𝑛 35°
d. 𝑐𝑜𝑠 7° e. 𝑡𝑎𝑛 85°