Trigonometry: Problems with Solutions

Solution:
$\sin \theta=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{4}{5}$

$\cos \theta=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{3}{5}$

$\tan \theta=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{4}{3}$

$\cot \theta=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{3}{4}$

Solution:
$\sin \theta=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{5}{13}$

$\cos \theta=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{12}{13}$

$\tan \theta=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{5}{12}$

$\cot \theta=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{12}{5}$

Solution:
The the Pythagorean Theorem states that $c^{2}=a^{2}+b^{2}$
Then $b=\sqrt{c^{2}-a^{2}}$
So $b=\sqrt{9-5}=\sqrt{4}=2$

$\sin \theta =\frac{b}{c}=\frac{2}{3}$

Solution:
Since $\cos 30^{\circ }=\frac{a}{c}\Longrightarrow c=\frac{a}{\cos 30^{\circ }}=\frac{2}{\frac{\sqrt{3}}{2}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$

According to the Pythagorean Theorem $c^{2}=a^{2}+b^{2}$

So $b=\sqrt{c^{2}-a^{2}}=\sqrt{\left( \frac{4\sqrt{3}}{3}\right) ^{2}-2^{2}}=\sqrt{\frac{16}{3}-4}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

Solution:
Since $\sin 30^{\circ }=\frac{b}{c}\Longrightarrow c=\frac{b}{\sin 30^{\circ}}=\frac{2}{\frac{1}{2}}=4$

According to the Pythagorean Theorem $c^{2}=a^{2}+b^{2}$

So $a=\sqrt{c^{2}-b^{2}}=\sqrt{4^{2}-2^{2}}=\sqrt{12}=2\sqrt{3}$

Solution:
We need to find the opposite side (os).
The Pythagorean Theorem says:
$\left( os\right) ^{2}+\left( as\right)^{2}=h^{2}\Longrightarrow \left(os\right) =\sqrt{h^{2}-\left(as\right)^{2}}$

where $os$: opposite side $as$: adjacent side and $h$: hypotenuse

$\left( os\right) =\sqrt{6^{2}-\left( 3\right) ^{2}}=\sqrt{36-9}=\sqrt{27}$

and now
$\sin \theta =\frac{o.s}{h}=\frac{\sqrt{27}}{6}$

Solution:
Let $h$ be the distance between the kite and the ground.
In the figure we see that $\frac{h}{90}=\sin 22^{0}$ and therefore $h=90\sin 22^{0}\approx 33.71$ ft.

Solution:
$\tan \theta =\frac{\sin \theta}{\cos \theta}=\frac{2}{3}$ and $\cot \theta =\frac{\cos \theta}{\sin \theta}=\frac{3}{2}$

Solution:
Let $x$, $y$, and $z$ be the unknown dimensions.
So, according to the definition of the tangent function,

$\tan 25^{0}=\frac{x}{4}$, $x=4\tan 25^{0}\backsimeq 1.87$ in

Let we calculate $y$
$\cos 25^{0}=\frac{4}{y}$, therefore $y=\frac{4}{\cos 25^{0}}\backsimeq 4.41$ in

We know that $z=\frac{3}{2}+x$ and $x\backsimeq 1.87$ in. So $z=1.5+1.87\backsimeq 3.37$ in.

Solution:
Since $\sin \theta =\frac{1}{\sqrt{65}}\Longrightarrow \text{opposite side}=1$

$\text{hypotenuse}=\sqrt{65}$ and
$\tan \theta =\frac{1}{8}\Longrightarrow \text{adjacent side}=8$

Then $\cos \theta =\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{8}{\sqrt{65}}=\frac{8\sqrt{65}}{65}$

and $\cot \theta =\frac{\text{adjacent side}}{\text{opposite side}}=8$

Solution:
We use the following trigonometric identity

$\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta -\sin \beta \cos \alpha $

So $\sin 15^{\circ }=\sin \left( 90^{\circ }-75^{\circ }\right) =\sin 90^{\circ}\cos 75^{\circ }-\sin 75^{\circ}\cos 90^{\circ }$

Since $\sin 90^{\circ }=1$ and $\cos 90^{\circ }=0$ we get

$\sin 15^{\circ }=\cos 75^{\circ }=\frac{1}{4}\left( \sqrt{6}-\sqrt{2}\right)$

Solution:
Having in mind the following formulas:
$\sin 2x=2\sin x\cos x$, $\cos 2x=\cos^{2}x-\sin^{2}x$

and $\cos^{2}x+\sin^{2}x=1$ we get

$\frac{\sin 2x}{1+\cos 2x}=\frac{2\sin x\cos x}{1+\cos^{2}x-\sin^{2}x}=\frac{2\sin x\cos x}{\cos^{2}x+\sin^{2}x+\cos^{2}x-\sin^{2}x}=\frac{2\sin x\cos x}{2\cos^{2}x}=\frac{\sin x}{\cos x}=\tan x$

Solution:
$1-\sin ^{4}\alpha =\left[ 1-\sin ^{2}\alpha \right] \left[ 1+\sin^{2}\alpha \right] $
and since $1-\sin ^{2}\alpha =\cos ^{2}\alpha$

We get $\frac{1-\sin^{4}\alpha }{\cos^{2}\alpha }=\frac{\left[ 1-\sin^{2}\alpha \right] \left[ 1+\sin ^{2}\alpha \right] }{\cos ^{2}\alpha }=\frac{\cos^{2}\alpha \left[ 1+\sin ^{2}\alpha \right] }{\cos ^{2}\alpha }=1+\sin ^{2}\alpha $

Since $\sin ^{2}\alpha =1-\cos ^{2}\alpha $,
we get $\frac{1-\sin^{4}\alpha }{\cos ^{2}\alpha }=1+1-\cos ^{2}\alpha$

So $\frac{1-\sin ^{4}\alpha }{\cos ^{2}\alpha }=2-\cos ^{2}\alpha $

Solution:
We will use, that $\cos 2\alpha =\cos^{2}\alpha -\sin ^{2}\alpha $ and $\sin 2\alpha =2\sin \alpha \cos \alpha$

$\frac{2\sin \alpha }{\tan 2\alpha }=\frac{2\sin \alpha \cos 2\alpha }{\sin2\alpha }=\frac{2\sin \alpha \left( \cos ^{2}\alpha -\sin ^{2}\alpha \right)}{2\sin \alpha \cos \alpha }=\frac{\left( \cos ^{2}\alpha -\sin ^{2}\alpha \right) }{\cos \alpha }=\cos \alpha -\frac{\sin ^{2}\alpha }{\cos \alpha }$

Solution:
We see that $\sin^{2}\alpha +\cos^{2}\alpha =1\Longrightarrow \cos^{2}\alpha =1-\sin ^{2}\alpha $ So

$\cos^{2}\alpha =\left( 1-\sin \alpha \right) \left( 1+\sin \alpha \right) \Longrightarrow \frac{\cos \alpha }{\left( 1+\sin \alpha \right) }=\frac{\left( 1-\sin \alpha \right) }{\cos \alpha }$

Solution:
$\sin^{2}\alpha \cos^{2}\alpha +\cos^{4}\alpha =\cos^{2}\alpha \left( \sin^{2}\alpha +\cos^{2}\alpha \right) $

Since $\sin^{2}\alpha +\cos^{2}\alpha =1$

We get $\sin ^{2}\alpha \cos^{2}\alpha +\cos ^{4}\alpha =\cos ^{2}\alpha $

Solution:
Since $\tan \alpha =\frac{h}{x}$ and $\tan \beta =\frac{h}{c-x}$ we express $x$ in terms of $h$
$x=\frac{h}{\tan \alpha }$ and now we substitute into the second equation $c-x=\frac{h}{\tan \beta }$

So $c-\frac{h}{\tan \alpha }=\frac{h}{\tan \beta }\Longrightarrow c=h\left( \frac{1}{\tan \beta }+\frac{h}{\tan \alpha }\right) =h\left( \frac{\tan \alpha +\tan \beta }{\tan \alpha \tan \beta }\right) $

Then $h=c\left( \frac{\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }\right)$

Solution:
Let $h$ be the height of the mountain. The figure shows that there are two right triangles that share a common side, $h$.

We get two equations with two unknowns, $h$ and $z$:
$\frac{h}{z+0.5}=\tan 37^{\circ}$ and $\frac{h}{z}\tan 41^{\circ}$

Evaluate $h$:

$h=\left( z+0.5\right) \tan 37^{\circ}$ and $h=z\tan 41^{\circ}$
The last two results are equal, and we reach an equation where we can determine the distance $z$:

$\left( z+0.5\right) \tan 37^{\circ}=z\tan 41^{\circ}$

$z=\frac{-0.5\tan 37^{\circ}}{\tan 37^{\circ}-\tan 41^{\circ}}$

Now we can calculate $h$, knowing that $h=z\tan 41^{\circ}$

$h=\frac{-0.5\tan 37^{\circ}\tan 41^{\circ}}{\tan 37^{\circ}-\tan 41^{\circ}}\backsimeq 2.83$ km