Trigonometry – Tangent – Lengths – 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 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 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 (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 (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) 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

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)

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

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

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

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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