Right triangle trig from two ratios Practice Questions

Right triangle trig from two ratios. Game on.

Right triangle trig from two ratios is a grade 10 math skill aligned to Common Core standard HSG.SRT.C.6: understand that side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 right triangle trig from two ratios problems our math games drill.

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Warm-upeasy

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

$\cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} = \dfrac{\dfrac{5}{13}}{\dfrac{5}{12}} = \dfrac{5}{13} \cdot \dfrac{12}{5} = \dfrac{12}{13}$ , or from the $5$ - $12$ - $13$ triangle: adjacent $12$ , hypotenuse $13$ .

Mid-gameeasy

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

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\dfrac{7}{25}}{\dfrac{24}{25}} = \dfrac{7}{25} \cdot \dfrac{25}{24} = \dfrac{7}{24}$ .

Mid-gameeasy

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

$\cos\theta = \dfrac{\sin\theta}{\tan\theta} = \dfrac{\dfrac{12}{13}}{\dfrac{12}{5}} = \dfrac{12}{13} \cdot \dfrac{5}{12} = \dfrac{5}{13}$ .

Mid-gameeasy

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

$\sin\alpha = \tan\alpha \cdot \cos\alpha = \dfrac{12}{5} \cdot \dfrac{5}{13} = \dfrac{12}{13}$ .

Mid-gameeasy

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

$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\dfrac{15}{17}}{\dfrac{8}{17}} = \dfrac{15}{17} \cdot \dfrac{17}{8} = \dfrac{15}{8}$ .

Mid-gameeasy

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

$\cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} = \dfrac{\dfrac{4}{5}}{\dfrac{4}{3}} = \dfrac{4}{5} \cdot \dfrac{3}{4} = \dfrac{3}{5}$ .

Mid-gameeasy

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

$\sin\theta = \tan\theta \cdot \cos\theta = \dfrac{5}{12} \cdot \dfrac{12}{13} = \dfrac{5}{13}$ .

Buzzer beatereasy

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

$\sin\alpha = \tan\alpha \cdot \cos\alpha = \dfrac{7}{24} \cdot \dfrac{24}{25} = \dfrac{7}{25}$ .

Right triangle trig from two ratios. Game on.

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