1. Trigonometry – Tangent – Demonstration
This resource provides animated demonstrations of the mathematical method.
Check animations and delete slides not needed for your class.

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. TOA 𝑥 (O) 52° 12 cm (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑇𝑎𝑛 52×12=𝑂
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(O)
We want to find O.
(so cover O) 𝑥
𝑇𝑎𝑛 θ×𝐴 =𝑂
52°
𝑇𝑎𝑛 52×12=𝑂
=15.36 cm
12 cm
(A)

4. TOA 𝑥 (O) (A) 7 cm 54° 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑇𝑎𝑛 54×7=𝑂 =9.63
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(A)
(O) 𝑥
7 cm
We want to find O.
(so cover O)
54°
𝑇𝑎𝑛 θ×𝐴 =𝑂
𝑇𝑎𝑛 54×7=𝑂
=9.63 cm

5. TOA 𝑥 (O) 4 cm 8 cm (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ= 𝑂 𝐴 𝑇𝑎𝑛 𝑥= 4 8
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(O)
4 cm
We want to find Tan θ.
(so cover Tan θ) 𝑥
𝑇𝑎𝑛 θ= 𝑂 𝐴
𝑇𝑎𝑛 𝑥= 4 8
8 cm
(A)
𝑇𝑎𝑛 − 𝑥=
=26.57°

6. TOA 𝑥 (O) 12 cm 64° (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑂 𝑇𝑎𝑛 θ =𝐻 12 𝑇𝑎𝑛 64 =𝐻
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(O)
12 cm
We want to find A.
(so cover A)
𝑂 𝑇𝑎𝑛 θ =𝐻
64° 𝑥
(A)
12 𝑇𝑎𝑛 64 =𝐻
=5.85 cm

7. TOA 𝑥 5 cm (O) (A) 29° 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑂 𝑇𝑎𝑛 θ =𝐻 5 𝑇𝑎𝑛 29 =𝐻 =9.02
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
5 cm
(O)
(A) 𝑥
We want to find A.
(so cover A)
𝑂 𝑇𝑎𝑛 θ
29° =𝐻
5 𝑇𝑎𝑛 29 =𝐻
=9.02 cm

8. TOA 𝑥 (A) 10 cm 11 cm (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ= 𝑂 𝐴 𝑇𝑎𝑛 𝑥= 11 10
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(A)
10 cm 𝑥
We want to find Tan θ.
(so cover Tan θ)
𝑇𝑎𝑛 θ= 𝑂 𝐴
𝑇𝑎𝑛 𝑥= 11 10
11 cm
(O)
𝑇𝑎𝑛 − 𝑥=
=47.73°

9. TOA 𝑥 (A) 13 cm 8 cm (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ= 𝑂 𝐴 𝑇𝑎𝑛 𝑥= 8 13
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A 𝑥
(A)
We want to find Tan θ.
(so cover Tan θ)
13 cm
𝑇𝑎𝑛 θ= 𝑂 𝐴
𝑇𝑎𝑛 𝑥= 8 13
𝑇𝑎𝑛 − 𝑥=
=31.61°
8 cm
(O)

10. TOA 𝑥 (O) 4 cm 25° (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑂 𝑇𝑎𝑛 θ =𝐻 4 𝑇𝑎𝑛 25 =𝐻 =8.58
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(O)
4 cm
We want to find A.
(so cover A)
25°
𝑂 𝑇𝑎𝑛 θ 𝑥 =𝐻
(A)
4 𝑇𝑎𝑛 25 =𝐻
=8.58 cm

11. TOA 𝑥 (A) 11 cm 20 cm (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ= 𝑂 𝐴 𝑇𝑎𝑛 𝑥= 20 11
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(A)
11 cm 𝑥
We want to find Tan θ.
(so cover Tan θ)
𝑇𝑎𝑛 θ= 𝑂 𝐴
𝑇𝑎𝑛 𝑥= 20 11
20 cm
(O)
𝑇𝑎𝑛 − 𝑥=
=61.19°

12. TOA 𝑥 32° 9 cm (O) (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑇𝑎𝑛 32×9=𝑂 =5.62
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
32°
We want to find O.
(so cover O) 𝑥
9 cm
(O)
𝑇𝑎𝑛 θ×𝐴 =𝑂
(A)
𝑇𝑎𝑛 32×9=𝑂
=5.62 cm

13. TOA 𝑥 14 cm (O) 3 cm (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ= 𝑂 𝐴 𝑇𝑎𝑛 𝑥= 14 3
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
We want to find Tan θ.
(so cover Tan θ)
14 cm
𝑇𝑎𝑛 θ= 𝑂 𝐴
𝑇𝑎𝑛 𝑥= 14 3 𝑥
(O)
𝑇𝑎𝑛 −
3 cm 𝑥=
=77.91°
(A)

14. TOA 𝑥 𝑥 𝑥 11 cm 4 cm 46° 9 cm 6 cm 68° 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Calculate 𝑥 for these three triangles. (2dp) O
Tan θ A
11 cm
4 cm
46°
9 cm
6 cm 𝑥
68° 𝑥 𝑥

15. TOA 𝑥 𝑥 𝑥 11 cm 4 cm 46° 9 cm 6 cm 68° 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O 𝑥=4.44 𝑐𝑚 A
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Calculate 𝑥 for these three triangles. (2dp) O
Tan θ A
𝑥=4.44 𝑐𝑚
𝑥=6.21 𝑐𝑚
11 cm
4 cm
46°
9 cm
6 cm 𝑥
68° 𝑥
𝑥=23.96° 𝑥

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