1. Trigonometry – Tangent – Lengths – 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 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

6. 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

7. 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

8. 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

9. TOA 𝑥 (A) 10 cm 46° (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑇𝑎𝑛 46×10=𝑂
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
46°
We want to find O.
(so cover O) 𝑥
𝑇𝑎𝑛 θ×𝐴 =𝑂
(O)
𝑇𝑎𝑛 46×10=𝑂
=10.36 cm

10. TOA 𝑥 37° (A) 8 cm (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑂 𝑇𝑎𝑛 θ =𝐻 8 𝑇𝑎𝑛 37 =𝐻
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
37°
(A)
We want to find A.
(so cover A) 𝑥
𝑂 𝑇𝑎𝑛 θ =𝐻
8 𝑇𝑎𝑛 37 =𝐻
8 cm
=10.62 cm
(O)

11. TOA 𝑥 (A) 8 cm 59° (O) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑇𝑎𝑛 59×8=𝑂 =13.31
Label the sides.
Write the formula.
Substitute & calculate.
𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗
Find the value of 𝑥 to 2dp. O
Tan θ A
(A)
8 cm
59°
We want to find O.
(so cover O)
𝑇𝑎𝑛 θ×𝐴 =𝑂 𝑥
(O)
𝑇𝑎𝑛 59×8=𝑂
=13.31 cm

12. TOA 𝑥 14 cm (O) 68° (A) 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O A 𝑂 𝑇𝑎𝑛 θ =𝐻 14 𝑇𝑎𝑛 68 =𝐻
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 A.
(so cover A)
14 cm
𝑂 𝑇𝑎𝑛 θ =𝐻
(O)
68° 𝑥
14 𝑇𝑎𝑛 68 =𝐻
(A)
=5.66 cm

13. TOA 𝑥 𝑥 𝑥 57° 12 cm 18° 5 cm 6 cm 29° 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 𝑥
57°
12 cm
18°
5 cm
6 cm 𝑥
29° 𝑥

14. TOA 𝑥 𝑥 𝑥 57° 12 cm 18° 5 cm 6 cm 29° 𝑇𝑎𝑛 θ= 𝑂𝑝𝑝 𝐴𝑑𝑗 O 𝑥=7.70 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=7.70 𝑐𝑚 𝑥
57°
12 cm
𝑥=10.82 𝑐𝑚
18°
5 cm
6 cm 𝑥
29° 𝑥
𝑥=3.90 𝑐𝑚

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