Exercises: Sine, Cosine and Tangent on a right triangle

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Calculate the missing sides and angles of the isosceles triangle $ABC$ with $a=b$ . Note that we are looking at general isosceles triangles that are not necessarily right-angled.

a=44.2cm

c=63.4cm

For this task you need the following basic knowledge: Sine, Cosine and Tangent

Given: a=b= 44.2cm c=63.4cm

Required: h, $\alpha,\;\beta,\;\gamma$

Draw a sketch for an overview.

Skizze des Dreiecks mit den zu bestimmenden Größen

First compute $x$ .

$\displaystyle x$	$=$	$\displaystyle \frac{c}{2}$
$\displaystyle x$	$=$	$\displaystyle \frac{63.4cm}{2}$
	$=$	$\displaystyle 31.7cm$

Compute $h$ by using the Pythagorean theorem within the triangle $\triangle DBC$ .

$\displaystyle h$	$=$	$\displaystyle \sqrt{a^2-x^2}$
	↓	Plug in known values.
	$=$	$\displaystyle \sqrt{\left(44.2cm\right)^2-\left(31.7cm\right)^2}$
	↓	square
	$=$	$\displaystyle \sqrt{1953.64cm^2-1004.89cm^2}$
	↓	subtract
	$=$	$\displaystyle \sqrt{948.75cm^2}$
	↓	take the root
	$=$	$\displaystyle 30.8cm$

Compute $\alpha$ usig the sine.

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h}{b}$
	↓	Plug in known values.
	$=$	$\displaystyle \frac{30.8cm}{44.2cm}$
	↓	Compute $\alpha$ using a calculator.
$\displaystyle \alpha$	$=$	$\displaystyle 44.2°\ =\ \beta$
	↓	So we have a right triangle with $\alpha=\beta$
	$=$

Since the base angles in an isosceles triangle are equal, and all interior angles total $180^\circ$ , you may calculate $\gamma$ .

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot44.2°$
	↓	multiply and subtract
	$=$	$\displaystyle 91.6°$

$\Rightarrow$ $h=30.8cm;\alpha=\beta=44.2^{\circ};\gamma=91.6^{\circ}$

Attention: The triangle $ABC$ is not a right triangle because no angle is $90°$ .

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a=114.5m

$\alpha$ =32.3°

For this task you need the following basic knowledge: Sine, Cosine and Tangent

Given: $a=b= 114{,}5\,m$ $\alpha=\beta$ =32.3°

Required: $c$ , $h$ , $\gamma$

Draw a sketch for an overview.

Since the base angles in an isosceles triangle are equal, and all interior angles total $180^\circ$ , you may calculate $\gamma$ .

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot32.3°$
	$=$	$\displaystyle 115.4°$

$h$ can be calculated using the sine.

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h}{b}$
	↓	solve for $h$ and plug in values.
$\displaystyle h$	$=$	$\displaystyle 114.5m\cdot\sin\left(32.3°\right)$
	↓	multiply
	$=$	$\displaystyle 61.2m$

$x$ is computed using the Pythagorean theorem within the triangle $\triangle DBC$ .

$\displaystyle x$	$=$	$\displaystyle \sqrt{a^2-h^2}$
	↓	Plug in known values.
	$=$	$\displaystyle \sqrt{\left(114.5m\right)^2-\left(61.2m\right)^2}$
	↓	square
	$=$	$\displaystyle \sqrt{13110.25m^2-3745.44m^2}$
	↓	subtract
	$=$	$\displaystyle \sqrt{9364.81m^2}$
	↓	take the root
$\displaystyle x$	$=$	$\displaystyle 96.8cm$

$c$ can be obtained by doubling the length of $x$ and then cutting the height $h$ of $c$ in half, such that $2$ equally long pieces $x$ are created.

$c=2\cdot96.8m=193.6m$

$\Rightarrow$ $h=61.2\,m; c=193.6\,m;\gamma=115.4^\circ$

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c=35.4cm

$\beta$ =43.9°

For this task you need the following basic knowledge: Sine, Cosine and Tangent

Given: c=35.4cm $\beta=\alpha$ =43.9°

Required: a, b, h, $\gamma$ , x

Draw a sketch for an overview.

https://assets.serlo.org/legacy/57d418ef6335c_393474c1c85c0622ebb9e53ff9cf2a021b3a1f64.png

Since the base angles in an isosceles triangle are equal, and all interior angles total $180^\circ$ , you may calculate $\gamma$ .

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot43.9°$
	$=$	$\displaystyle 92.2°$

x is computed by cutting c into half.

$\displaystyle x$	$=$	$\displaystyle \frac{35.4cm}{2}$
	$=$	$\displaystyle 17.7cm$

a can be calculated using the cosine.

$\displaystyle \cos\left(\beta\right)$	$=$	$\displaystyle \frac{x}{a}$
	↓	Solve for $a$ and plug in values.
$\displaystyle a$	$=$	$\displaystyle \frac{17.7cm}{\cos\left(43.9°\right)}$
	↓	divide
$\displaystyle a$	$=$	$\displaystyle 24.6cm$

h can be calculated using the tangent.

$\displaystyle \tan\left(\beta\right)$	$=$	$\displaystyle \frac{h}{x}$
	↓	Solve for $h$ and plug in values.
$\displaystyle h$	$=$	$\displaystyle 17.7cm\cdot\tan\left(43.9°\right)$
	↓	multiply
$\displaystyle h$	$=$	$\displaystyle 17.0cm$

$\Rightarrow\;\;$ $\alpha=43.9^\circ;\;\gamma=92.2^\circ;\;a=b=24.6cm;\;h=17.0cm\;$

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h=14.8cm

$\alpha=\beta=$ 28.3°

For this task you need the following basic knowledge: Sine, Cosine and Tangent

Given.: $h=14.8cm$ ; $\alpha=\beta=28.3^{\circ}$

Required.: $\beta,\gamma,c, b, a$

Draw a sketch for an overview.

Since the base angles (here: $\alpha$ and $\beta$ ) in an isosceles triangle are equal, and all the interior angles add up to $180^\circ$ (i.e. $\alpha+\beta+\gamma=180^\circ$ ), you can calculate $\gamma$ directly with this information.

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot28.3°$
	$=$	$\displaystyle 123.4°$

x can be calculated using the tangent.

$\displaystyle \tan\left(\beta\right)$	$=$	$\displaystyle \frac{h}{x}$
	↓	Solve for $x$ and plug in values.
$\displaystyle x$	$=$	$\displaystyle \frac{14.8cm}{\tan\left(28.3°\right)}$
	↓	divide
$\displaystyle x$	$=$	$\displaystyle 27.5cm$
	↓	You get $c$ by doubling side $x$ .
$\displaystyle c$	$=$	$\displaystyle 2\cdot27.5cm$
	$=$	$\displaystyle 55cm$

$b$ can be calculated using the sine.

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h}{b}$
	↓	Solve for $b$ and plug in values.
$\displaystyle b$	$=$	$\displaystyle \frac{14.8cm}{\sin\left(28.3°\right)}$
	↓	divide
	$=$	$\displaystyle 31.2cm$

Since this is an isosceles triangle, the side length $a$ is just equal to the side length $b$ .

$\Rightarrow\;\;$ $\gamma=123.4^\circ;\;c=55cm;\;a=b=31.2cm$

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a=146.4m

h=58.4m

For this task you need the following basic knowledge: Sine, Cosine and Tangent

Given: $a=b=146.4 \, m$ ; $h=58.4\, m$

Required: $c$ , $\gamma,\;\alpha,\;\beta$ , $x$

Draw a sketch for an overview.

$\beta$ can be calculated using the sine.

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h}{a}$
	↓	Plug in values and compute $\alpha$
$\displaystyle \beta$	$=$	$\displaystyle 23.5°$

Since the base angles in an isosceles triangle are equal, and all interior angles total $180^\circ$ , you may calculate $\gamma$ .

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot23.5°$
	$=$	$\displaystyle 133°$

$x$ can be calculated using the tangent.

$\displaystyle \tan\left(\beta\right)$	$=$	$\displaystyle \frac{h}{x}$
	↓	Solve for $x$ and plug in values.
$\displaystyle x$	$=$	$\displaystyle \frac{58.4m}{\tan\left(23.5°\right)}$
	$=$	$\displaystyle 134.3m$
	↓	$c$ is computed by doubling side $x$ and then cutting the height $h$ of $c$ into half, such that $2$ equally long pieces $x$ result.
$\displaystyle c$	$=$	$\displaystyle 2\cdot134.3m$
	$=$	$\displaystyle 268.6m$

$\Rightarrow\;\;$ $b=146.4m;\;\alpha=\beta=23.5^\circ;\;\gamma=133^\circ;\;c=268.5m$

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