Find perpendicular height of triangle Consider a triangle $\triangle ABC$. It is given that angle $A\hat{B}C$ is 38 degrees, angle $A\hat{C}B$ is 62 degrees, the length of $\overline{BC}$ is 38 cm. Find the perpendicular height of the triangle (from A to the base BC).
I can solve this with the use of Sine Rule. But is there another way to solve this question without using Sine Rule?
 A: Let the perpendicular from A onto the side BC meets BC at the point D.
Let the length BD be $x$ cm. Therefore, CD becomes $(38-x)$ cm.
Now, tangent of angle ABC (which is 38 degrees) = $AD/BD$ = $AD/x$ ....... (1)
Also, tangent of angle ACB (which is 62 degrees) = $AD/CD$ = $AD/(38-x)$ ..... (2)
Now, equate the value of AD from equation (1) and equation (2) to get the value of x as $26.85$ cm, approximately. Put this back in equation (1) to get $AD = 20.98$ cm, approximately, which is what you wanted!
A: 

$\angle ABC=\beta=38^\circ$,  $\angle ACB=\gamma=62^\circ$, 
  $|BC|=a=38$cm, $D\in BC,\ AD\perp BC$  Find $|AD|=h_a$

Known formula for the area of triangle
in terms of given side length and two adjacent angles
is
\begin{align}
S&=\tfrac12\,a^2\,\frac{\sin\beta\sin\gamma}
{\sin(\beta+\gamma)}
,
\end{align}
on the other hand,
\begin{align}
S&=\tfrac12\,a\,h_a
,\\
\text{so }\quad
h_a&=a\,\frac{\sin\beta\sin\gamma}{\sin(\beta+\gamma)}
.
\end{align}
A: 
No sine(s) used:
\begin{align}
\triangle{ABD}:\quad
\tan\beta&=\frac h{|BD|}
,\quad \Rightarrow |BD|=h\,\cot\beta
,\\
\triangle{ADC}:\quad
\tan\gamma&=\frac h{|CD|}
,\quad \Rightarrow |CD|=h\,\cot\gamma
,\\
|BD|+|CD|&=|BC|
,\\
h\,\cot\beta+h\,\cot\gamma
&=|BC|
,\\
h&=\frac{|BC|}{\cot\beta+\cot\gamma}
.
\end{align}
