trigonometry
A bridge is built across the top of two buildings of the same height. The length of the bridge is 10 m. If from a point on the ground between the two buildings, the angles of elevation of the ends of the bridge are 68 degrees and 62 degrees, how high are the buildings?
Can i solve this using (law of Sine) and ( law of Cosine) ?
Yes, this uniquely defines the sine and cosine functions. In particular, let's write $$f(x)=\cos(x)+i\sin(x)$$ where $i$ is the imaginary unit. Then, the sum identities will yield, after a bit of computation that $f(x)f(y)=f(x+y)$. This is somewhat tedious, but easy to verify. But guess what! The only continuous functions which can satisfy $f(x)f(y) = f(x+y)$ are exponential functions; to prove this, notice that, for integer $n$, it is clear that $f(nx)=f(x)^n$. You can use this to show that, over the rationals, $f$ is an exponential function (i.e. $f(x)=e^{ax}$).*
Since we have $\sin(0)=0$, $\sin'(0)=1$, $\cos(0)=1$ and $\cos'(0)=0$, this implies that $f'(0)=i$. The only exponential function satisfying this is $f(x)=e^{ix}$. Extracting real and imaginary parts yields, uniquely, cosine and sine.
*If you wish to be formal about this, it would be wise to prove that $|f(x)|=e^{\alpha x}$ first, then prove that $\arg(f(x))\equiv \beta x$ - you can avoid the issue of $n^{th}$ roots being non-unique in the complex plane by separating your argument into these two sections.
The observer who measures the angle of elevation to the rocket to be $66^\circ$ is closer than the observer who measures the angle of elevation to be $37^\circ$.
Consequently, if $x$ is the distance from the closer observer to the point where the rocket is launched, then
$$\tan(66^\circ) = \frac{h}{x}$$
and, since the second observer is $20~\text{m}$ farther from the launch site,
$$\tan(37^\circ) = \frac{h}{x + 20~\text{m}}$$
Best Answer
Consider the following figure.
Points $B$ and $C$ are the endpoints of the bridge. Point $A$ is the point on the ground from which the angles of elevation are measured.
Observe that
\begin{align*} \tan(68^\circ) & = \frac{h}{x}\\ \tan(62^\circ) & = \frac{h}{10~\text{m} - x} \end{align*}
Solving the equation
$$\tan(68^\circ) = \frac{h}{x}$$
for $h$ yields
$$h = x\tan(68^\circ)$$
Solving the equation
$$\tan(62^\circ) = \frac{h}{10~\text{m} - x}$$
for $h$ yields
$$h = (10~\text{m} - x)\tan(62^\circ)$$
We can solve for $x$ by equating the two expressions for $h$.
\begin{align*} x\tan(68^\circ) & = (10~\text{m} - x)\tan(62^\circ)\\ x\tan(68^\circ) & = 10~\text{m}\tan(62^\circ) - x\tan(62^\circ)\\ x\tan(68^\circ) + x\tan(62^\circ) & = 10~\text{m}\tan(62^\circ)\\ x[\tan(68^\circ) + \tan(62^\circ)] & = 10~\text{m}\tan(62^\circ)\\ x & = \frac{10~\text{m}\tan(62^\circ)}{\tan(68^\circ) + \tan(62^\circ)} \end{align*}
Substituting for $x$ in the equation
$$h = x\tan(68^\circ)$$
yields
$$h = \frac{10~\text{m}\tan(62^\circ)\tan(68^\circ)}{\tan(68^\circ) + \tan(62^\circ)} \approx 10.69~\text{m}$$
which is about the height of a three-story building.