Trigonometry – 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.

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)

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 49 =𝐻 𝑥 =34.45 cm (H)

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)

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)

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

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)

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)

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

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

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

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 53 =𝐻 =17.53 cm 𝑥 (H)

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)

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

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

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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