Trigonometric ratios problem

Question

How can we find $x$ in this question without knowing $sin$, $cos$, $tan$, $cot$, $sec$ or $csc$ of given angles

Answer 1

$$x\cos15°\sin15°=\text{BA}\sin15°=3$$ so that $$x=\frac6{\sin30°}=12.$$

(Assuming that you are allowed to "know" the value of $\sin30°$. Otherwise observe in the figure that $\text{AB}=\text{OA}=2\text{AN}$.)

Answer 2

Since $$\begin{align}\frac{AH}{AB}&=\sin(15^{\circ})\\ \frac{AB}{BC}&=\cos(15^{\circ}) \end{align}$$ (why?) we have $$\frac{3}{x}=\frac{AH}{BC}=\sin(15^{\circ})\cos(15^{\circ})\text{.}$$ You really need to know something about $\sin(15^{\circ})\cos(15^{\circ})$—or perhaps $2\,\sin(15^{\circ})\cos(15^{\circ})$?—to proceed from here.

Answer 3

Extend $\overrightarrow{HA}$ to point $D$ such that $m\angle DBA = 15^\circ$

Let $HD = x$. Since $\triangle BDH$ is a $30-60-90$ right triangle, then $AD = x-3$, $BD = 2x$, $BX = \sqrt 3 x$

By the angle bisector theorem,

\begin{align} \dfrac{2x}{x-3} &= \dfrac{\sqrt 3x}{3} \\ \dfrac{x-3}{3} &= \dfrac{2}{\sqrt 3} \\ x-3 &= 2 \sqrt 3 \\ x &= 3+2\sqrt 3 \\ \hline AH &= 3 \\ AD &= 2\sqrt 3 \\ BD &= 6+4\sqrt 3 \\ BH &= 6 + 3\sqrt 3 \\ BA &= 6\sqrt{2+\sqrt 3} \end{align}

Since $\triangle ACH \cong \triangle BAH$, then \begin{align} \dfrac{AH}{CH} &= \dfrac{BH}{AH} \\ AH^2 &= BH \cdot CH \\ 9 &= (6+3\sqrt 3)CH \\ CH &= \dfrac{3}{2+\sqrt 3} \\ CH &= 6-3\sqrt 3 \\ \hline BC &= 12 \end{align}

Answer 4

We know that in a triangle with angles $30,60,90$ degrees, the opposed side to $30$ is half of hypotenuse. This statement no needs trigonometry.

Now using this, try to prove that in the given triangle " the altitude is $1/4$ of hypotenuse" by drawing the median $AM$.