I have problem with Trigonometry

Question

Tomorrow I have a test and there is one exercise in my textbook that isn't explained. Here is the exercise.

tangens(tg) from alfa = 4/3. 
Find : sinus(sin) from alfa and co-tangens(cotg) from alfa.

I will be very grateful if somebody explain how to solve those kind of exercises.

Answer 1

We have $$ \tan\alpha=\frac yx=\frac43, $$ then $$ \sin\alpha=\frac y{\sqrt{x^2+y^2}}=\frac 4{\sqrt{3^2+4^2}}=\frac45 $$ and $$ \cot\alpha=\frac xy=\frac1{\tan\alpha}=\frac1{\frac43}=\frac34. $$

Answer 2

You can do this by drawing a right angle triangle with adjacent side 3 and opposite side 4. Since: $$\tan(\alpha)=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{3}$$ We can then work out the hypotenuse:$$ \sqrt{(3^2+4^2)}=5 $$

Therefore given the definition of sine as the opposite side divided by the hypotenuse thus gives: $$\sin({\alpha)}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{4}{5}$$ and the cotangent is the reciprocal of the tangent which we already know is $\frac{4}{3}$ therefore: $$ \cot{(\alpha)}=\frac{1}{\tan(\alpha)}=\frac{1}{\frac{4}{3}}=\frac{3}{4} $$

Answer 3

So you were given the value of $\tan(\alpha)$ and you were asked to find $\sin(\alpha)$ and $\cot(\alpha)$. To do this, we must be able to know some relations relating $\sin(\alpha)$ to $\tan(\alpha)$ and $\cot(\alpha)$ to $\tan(\alpha)$, one of those relations is: $$\cot(\alpha)\equiv \dfrac1{\tan(\alpha)}\Rightarrow\cot(\alpha)=\dfrac1{\frac43}=\dfrac34.$$ $$\sin(\alpha)\equiv\dfrac{b^2}{\sqrt{a^2+b^2}}\,\text{ given: }\tan{(\alpha)}=\dfrac ba\\ \Rightarrow \sin(\alpha)=\dfrac4{\displaystyle\sqrt{\displaystyle3^2+4^2}}=\dfrac4{\sqrt{25}}=\dfrac45.$$