1. Trigonometry – Lengths – Demonstration
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2. θ Hypotenuse Opposite Adjacent A right-angled triangle has 4 parts.
θ = Theta is either angle.
Hypotenuse
Opposite θ
Adjacent
Hypotenuse – always opposite the right-angle & always longest.
Opposite – always opposite θ.
Adjacent – next to θ.

3. SOH CAH TOA 𝑥 (H) 7 cm 62° (O) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find O.
(so cover O)
(H)
7 cm
𝑆𝑖𝑛 θ ×𝐻 =𝑂
62° 𝑥
𝑆𝑖𝑛 62×7=𝑂
=6.18 cm
(O)

4. SOH CAH TOA 𝑥 26 cm (O) 49° (H) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find H.
(so cover H)
26 cm
𝑂 𝑆𝑖𝑛 θ
(O) =𝐻
49°
26 𝑆𝑖𝑛 =𝐻 𝑥
=34.45 cm
(H)

5. SOH CAH TOA 𝑥 (H) 34° 12 cm (A) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find H.
(so cover H)
(H) 𝑥
𝐴 𝐶𝑜𝑠 θ =𝐻
34°
12 𝐶𝑜𝑠 34 =𝐻
=14.47 cm
12 cm
(A)

6. SOH CAH TOA 𝑥 (H) 13 cm 37° (A) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find A.
(so cover A)
(H)
13 cm
𝐶𝑜𝑠 θ×𝐻 =𝐴
37°
𝐶𝑜𝑠 37×13=𝐴
=10.38 cm 𝑥
(A)

7. SOH CAH TOA 𝑥 (O) (A) 4 cm 52° 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find O.
(so cover O) 𝑥
(O)
𝑇𝑎𝑛 θ×𝐴 =𝑂
(A)
4 cm
52°
𝑇𝑎𝑛 52×4=𝑂
=5.12 cm

8. SOH CAH TOA 𝑥 (O) 11 cm 34° (A) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find A.
(so cover A)
(O)
𝑂 𝑇𝑎𝑛 θ =𝐴
11 cm
11 𝑇𝑎𝑛 34 =𝐴
34°
=16.31 cm 𝑥
(A)

9. SOH CAH TOA 𝑥 (O) 41° 13 cm (H) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find O.
(so cover O)
(O) 𝑥
𝑆𝑖𝑛 θ×𝐻 =𝑂
41°
13 cm
𝑆𝑖𝑛 41×13=𝑂
=8.53 cm
(H)

10. SOH CAH TOA 𝑥 (A) 4 cm 62° (H) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find H.
(so cover H)
𝐴 𝐶𝑜𝑠 θ
(A) =𝐻
4 cm
62° 𝑥
(H)
4 𝐶𝑜𝑠 62 =𝐻
=8.52 cm

11. SOH CAH TOA 𝑥 28° (O) (A) 32 cm 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find A.
(so cover A)
𝑂 𝑇𝑎𝑛 θ
28° =𝐴
(O) 𝑥
(A)
32 cm
32 𝑇𝑎𝑛 28 =𝐴
=60.18 cm

12. SOH CAH TOA 𝑥 (A) 43° 25 cm (H) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find A.
(so cover A)
(A) 𝑥
𝐶𝑜𝑠 θ×𝐻 =𝐴
43°
25 cm
(H)
𝐶𝑜𝑠 43×25=𝐴
=18.28 cm

13. SOH CAH TOA 𝑥 14 cm (O) 53° (H) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find H.
(so cover H)
14 cm
𝑂 𝑆𝑖𝑛 θ
(O) =𝐻
53°
14 𝑆𝑖𝑛 =𝐻
=17.53 cm 𝑥
(H)

14. SOH CAH TOA 𝑥 (A) 57° 9 cm (O) 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Find the value of 𝑥 to 2dp.
We want to find O.
(so cover O)
(A)
𝑇𝑎𝑛 θ×𝐴 =𝑂
57°
9 cm
𝑇𝑎𝑛 57×9=𝑂
=13.86 cm 𝑥
(O)

15. SOH CAH TOA 𝑥 𝑥 𝑥 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O H A H
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Calculate 𝑥 for these three triangles. (2dp) 𝑥
28°
6 cm 𝑥
44°
6 cm
8 cm
15° 𝑥

16. SOH CAH TOA 𝑥 𝑥 𝑥 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O H A H
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Calculate 𝑥 for these three triangles. (2dp) 𝑥
28°
𝑇𝑂𝐴
𝑆𝑂𝐻
6 cm 𝑥
44°
6 cm
𝐶𝐴𝐻
8 cm
15° 𝑥

17. SOH CAH TOA 𝑥 𝑥 𝑥 𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O H A H
𝑆𝑖𝑛 θ= 𝑂𝑝𝑝 𝐻𝑦𝑝
𝐶𝑜𝑠 θ= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O
Sin θ H A
Cos θ H O
Tan θ A
Calculate 𝑥 for these three triangles. (2dp) 𝑥
𝑥=6.21 𝑐𝑚
28°
𝑇𝑂𝐴
𝑆𝑂𝐻
6 cm 𝑥
44°
6 cm
𝐶𝐴𝐻
8 cm
15° 𝑥
𝑥=12.78 𝑐𝑚
𝑥=7.73 𝑐𝑚

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