HSG.SRT.C.6 math practice. Learn by doing.

HSG.SRT.C.6 practice covers understand that side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Work through 8 free questions with answers and explanations, then continue in the related math games.

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In a right triangle, sin⁡α=513\sin\alpha = \dfrac{5}{13} and tan⁡α=512\tan\alpha = \dfrac{5}{12}. What is cos⁡α\cos\alpha?

cos⁡α=sin⁡αtan⁡α=513512=513⋅125=1213\cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} = \dfrac{\dfrac{5}{13}}{\dfrac{5}{12}} = \dfrac{5}{13} \cdot \dfrac{12}{5} = \dfrac{12}{13}, or from the 55-1212-1313 triangle: adjacent 1212, hypotenuse 1313.

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In a right triangle, sin⁡α=725\sin\alpha = \dfrac{7}{25} and cos⁡α=2425\cos\alpha = \dfrac{24}{25}. What is tan⁡α\tan\alpha?

tan⁡α=sin⁡αcos⁡α=7252425=725⋅2524=724\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\dfrac{7}{25}}{\dfrac{24}{25}} = \dfrac{7}{25} \cdot \dfrac{25}{24} = \dfrac{7}{24}.

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In a right triangle, sin⁡θ=1213\sin\theta = \dfrac{12}{13} and tan⁡θ=125\tan\theta = \dfrac{12}{5}. What is cos⁡θ\cos\theta?

cos⁡θ=sin⁡θtan⁡θ=1213125=1213⋅512=513\cos\theta = \dfrac{\sin\theta}{\tan\theta} = \dfrac{\dfrac{12}{13}}{\dfrac{12}{5}} = \dfrac{12}{13} \cdot \dfrac{5}{12} = \dfrac{5}{13}.

Practice playeasy

In a right triangle, cos⁡α=513\cos\alpha = \dfrac{5}{13} and tan⁡α=125\tan\alpha = \dfrac{12}{5}. What is sin⁡α\sin\alpha?

sin⁡α=tan⁡α⋅cos⁡α=125⋅513=1213\sin\alpha = \tan\alpha \cdot \cos\alpha = \dfrac{12}{5} \cdot \dfrac{5}{13} = \dfrac{12}{13}.

Practice playeasy

In a right triangle, sin⁡θ=1517\sin\theta = \dfrac{15}{17} and cos⁡θ=817\cos\theta = \dfrac{8}{17}. What is tan⁡θ\tan\theta?

tan⁡θ=sin⁡θcos⁡θ=1517817=1517⋅178=158\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\dfrac{15}{17}}{\dfrac{8}{17}} = \dfrac{15}{17} \cdot \dfrac{17}{8} = \dfrac{15}{8}.

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In a right triangle, sin⁡α=45\sin\alpha = \dfrac{4}{5} and tan⁡α=43\tan\alpha = \dfrac{4}{3}. What is cos⁡α\cos\alpha?

cos⁡α=sin⁡αtan⁡α=4543=45⋅34=35\cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} = \dfrac{\dfrac{4}{5}}{\dfrac{4}{3}} = \dfrac{4}{5} \cdot \dfrac{3}{4} = \dfrac{3}{5}.

Practice playeasy

In a right triangle, cos⁡θ=1213\cos\theta = \dfrac{12}{13} and tan⁡θ=512\tan\theta = \dfrac{5}{12}. What is sin⁡θ\sin\theta?

sin⁡θ=tan⁡θ⋅cos⁡θ=512⋅1213=513\sin\theta = \tan\theta \cdot \cos\theta = \dfrac{5}{12} \cdot \dfrac{12}{13} = \dfrac{5}{13}.

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In a right triangle, cos⁡α=2425\cos\alpha = \dfrac{24}{25} and tan⁡α=724\tan\alpha = \dfrac{7}{24}. What is sin⁡α\sin\alpha?

sin⁡α=tan⁡α⋅cos⁡α=724⋅2425=725\sin\alpha = \tan\alpha \cdot \cos\alpha = \dfrac{7}{24} \cdot \dfrac{24}{25} = \dfrac{7}{25}.

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