Exercises: Sine, Cosine and Tangent on a right triangle

Calculate the missing sides and angles (marked red) of the triangles

$γ = 90^{\circ}$
$a = 12.7 cm$
$c = 24.9 cm$

For this task you need the following basic knowledge: Trigonometry on the unit circle
computing $α$
$\sin (α) = \frac{12.7 cm}{24.9 cm}$
$α = {30.7}^{\circ}$
computing $β$
You can calculate $β$ by subtracting all given angles from the total sum of all angles in a triangle $(180^{\circ})$ .
$β = 180^{\circ} - 90^{\circ} - {30.7}^{\circ}$
$β = {59.3}^{\circ}$
computing $b$
Then you compute $b$ using the Pythagorean theorem.
${(24.9 cm)}^{2} = {(12.7 cm)}^{2} + b^{2}$
$| - {(12.7 cm)}^{2}$
$b^{2} = {(24.9 cm)}^{2} - {(12.7 cm)}^{2}$
$b^{2} = 458.72 {cm}^{2}$
Take the root.
$b \approx 21.4 cm$
$α = 90^{\circ}$
$b = 420 m$
$a = 645 m$

For this task you need the following basic knowledge: Trigonometry on the unit circle
computing $β$
$\sin (β) = \frac{420 m}{645 m}$
$β = {40.6}^{\circ}$
computing $γ$
You can calculate $γ$ by subtracting all given angles from the total sum of all angles in a triangle $(180^{\circ})$ .
$γ = 180^{\circ} - 90^{\circ} - {40.6}^{\circ} = {49.4}^{\circ}$
computing $c$
Then you compute $c$ using the Pythagorean theorem.
Note in particular that c is not the hypotenuse of the triangle, but $a$ (see picture above), so that the Pythagorean form remains similar, but now $a$ , $b$ and $c$ do not take the familiar roles, where $a$ , $b$ would be the cathetuses and $c$ the hypotenuse.
$(645 m)^{2} = (420 m)^{2} + c^{2}$
$c^{2} = (645 m)^{2} - (420 m)^{2}$
$c^{2} = 239 625 m^{2}$
Take the root.
$c \approx 490 m$
$β = 90^{\circ}$
$c = 15.8 cm$
$a = 30.7 cm$

For this task you need the following basic knowledge: Trigonometry on the unit circle
computing $b$
Compute $b$ using the Pythagorean theorem.
$b^{2} = {(30.7 cm)}^{2} + {(15.8 cm)}^{2}$
$b^{2} = 1192.13 {cm}^{2} | \sqrt{}$
$b \approx 34.5 cm$
computing $α$
$\tan (α) = \frac{a}{c}$
$\tan (α) = \frac{30.7 cm}{15.8 cm}$
$α \approx {62.7}^{\circ}$
computing $γ$
You may calculate $γ$ by subtracting all given angles from the total sum of all angles in a triangle $(180^{\circ})$ .
$γ \approx 180^{\circ} - 90^{\circ} - {62.7}^{\circ}$
$γ \approx {27.3}^{\circ}$
$γ = 90^{\circ}$
$α = 35^{\circ}$
$c = 12.5 cm$

For this task you need the following basic knowledge: Trigonometry on the unit circle
computing $β$
You may calculate $β$ by subtracting all given angles from the total sum of all angles in a triangle $(180^{\circ})$ .
$β = 180^{\circ} - 90^{\circ} - 35^{\circ}$
$β = 55^{\circ}$
computing $a$
$\sin (35^{\circ}) = \frac{a}{12.5 cm} | \cdot 12.5 cm$
$a = \sin (35^{\circ}) \cdot 12.5 cm$
$a = 7.2 cm$
computing $b$
Then you compute $b$ using the Pythagorean theorem.
$c^{2} = a^{2} + b^{2}$
${(12.5 cm)}^{2} = {(7.2 cm)}^{2} + b^{2} | - {(7.2 cm)}^{2}$
$b^{2} = {(12.5 cm)}^{2} - {(7.2 cm)}^{2}$
$b^{2} \approx 104.4 {cm}^{2}$
Take the root.
$b \approx 10.2 cm$
computing $b$ (alternative solution with the cosine)
$\cos (α) = \frac{b}{c}$
Solve for $b$
$b = \cos (α) \cdot c \approx 10.2 cm$
$α = 90^{\circ}$
$γ = {40.3}^{\circ}$
$a = 10.5 cm$

For this task you need the following basic knowledge: Trigonometry on the unit circle
computing $β$
You may calculate $β$ by subtracting all given angles from the total sum of all angles in a triangle $(180^{\circ})$ .
$β = 180^{\circ} - 90^{\circ} - {40.3}^{\circ}$
$β = {49.7}^{\circ}$
computing $b$
$\sin ({49.7}^{\circ}) = \frac{b}{10.5 cm} | \cdot 10.5 cm$
$b = \sin ({49.7}^{\circ}) \cdot 10.5 cm$
$b \approx 8 cm$
computing $c$
Then you compute $c$ using the Pythagorean theorem.
${(10.5 cm)}^{2} = c^{2} + {(8 cm)}^{2}$
$c^{2} = 46.25 {cm}^{2}$
Take the root.
$c \approx 6.8 cm$

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