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The altitude AD of $\Delta ABC$ in which $\angle A$ is obtuse and AD = 10cm. If BD = 10 cm and $CD=10\sqrt{3}cm$, determine $\angle A$.

Answer
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Hint: Calculate the value of $\angle CAD\ and\ \angle BAD$using trigonometric ratios like sine, cosine and tangent.

Complete step by step answer:




In $\Delta ACD$, since $\Delta ACD$ is a right angled triangle,
$\begin{align}
  & \tan \left( \angle CAD \right)=\dfrac{CD}{AD} \\
 & =\dfrac{10\sqrt{3}}{10} \\
 & =\sqrt{3} \\
 & \Rightarrow \angle CAD={{\tan }^{-1}}\sqrt{3} \\
 & \Rightarrow \angle CAD=\dfrac{\pi }{3} \\
\end{align}$
We will now find $\angle BAD$, since $\Delta ABD $ is a right angled triangle,
\[\begin{align}
  & \tan \left( \angle BAD \right)=\dfrac{BD}{AD} \\
 & =\dfrac{10}{10} \\
 & =1 \\
 & \Rightarrow \angle BAD={{\tan }^{-1}}1 \\
 & \Rightarrow \angle BAD=\dfrac{\pi }{4} \\
 & \angle A=\angle CAD+\angle BAD \\
 & =\dfrac{\pi }{3}+\dfrac{\pi }{4} \\
 & =\dfrac{7\pi }{12} \\
 & =105{}^\circ \\
\end{align}\]
Hence $\angle A$, as asked to us in the question = $\dfrac{7\pi }{12}\ or\ 105{}^\circ $.

Note: The question can also be solved using the cosine and sine formula after finding the length of hypotenuse of each triangle using Pythagoras Theorem.