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

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.

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)

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

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 −1 4 8 𝑥= =26.57°

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

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

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 −1 11 10 𝑥= =47.73°

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 −1 8 13 𝑥= =31.61° 8 cm (O)

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

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 −1 20 11 𝑥= =61.19°

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

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 −1 14 3 3 cm 𝑥= =77.91° (A)

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

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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