Answer
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Hint: First we will draw the figure and mark all the values that are given so that we can get a clear picture of what we need to find, then we will use the formula $\tan \alpha =\dfrac{height}{base}$ and substitute the value of height as h and find the value of base for the two triangles and then add them to get the final answer.
Complete step-by-step answer:
Let’s first draw the figure,
In the above figure,
AB = h = height of the light house
DC = distance between the two ships
To find the value DC we need to find the value of BD and BC,
Let’s first find the value of DB,
In triangle ABD, height = AB = h and base = DB
$\tan \beta =\dfrac{height}{base}$
Now substituting the value of height = h and base = DB we get,
$\begin{align}
& \tan \beta =\dfrac{h}{DB} \\
& DB=\dfrac{h}{\tan \beta }...........(1) \\
\end{align}$
Let’s find the value BC,
In triangle ABC, height = AB = h and base = BC
$\tan \alpha =\dfrac{height}{base}$
Now substituting the value of height = h and base = BC we get,
$\begin{align}
& \tan \alpha =\dfrac{h}{BC} \\
& BC=\dfrac{h}{\tan \alpha }...........(2) \\
\end{align}$
Now we know that,
DC = DB + BC
Substituting the values of DB and BC from equation (1) and (2) we get,
$\begin{align}
& DC=\dfrac{h}{\tan \alpha }+\dfrac{h}{\tan \beta } \\
& DC=\dfrac{h\left( \tan \alpha +\tan \beta \right)}{\tan \alpha \tan \beta } \\
\end{align}$
Hence Proved.
Note: To check if the statement that we have proved is true or not, one can put the values of all the variables and check if DC = $\dfrac{h\left( \tan \alpha +\tan \beta \right)}{\tan \alpha \tan \beta }$ is true or not. Students must be able to understand the meaning of the question to solve it correctly.
Complete step-by-step answer:
Let’s first draw the figure,
In the above figure,
AB = h = height of the light house
DC = distance between the two ships
To find the value DC we need to find the value of BD and BC,
Let’s first find the value of DB,
In triangle ABD, height = AB = h and base = DB
$\tan \beta =\dfrac{height}{base}$
Now substituting the value of height = h and base = DB we get,
$\begin{align}
& \tan \beta =\dfrac{h}{DB} \\
& DB=\dfrac{h}{\tan \beta }...........(1) \\
\end{align}$
Let’s find the value BC,
In triangle ABC, height = AB = h and base = BC
$\tan \alpha =\dfrac{height}{base}$
Now substituting the value of height = h and base = BC we get,
$\begin{align}
& \tan \alpha =\dfrac{h}{BC} \\
& BC=\dfrac{h}{\tan \alpha }...........(2) \\
\end{align}$
Now we know that,
DC = DB + BC
Substituting the values of DB and BC from equation (1) and (2) we get,
$\begin{align}
& DC=\dfrac{h}{\tan \alpha }+\dfrac{h}{\tan \beta } \\
& DC=\dfrac{h\left( \tan \alpha +\tan \beta \right)}{\tan \alpha \tan \beta } \\
\end{align}$
Hence Proved.
Note: To check if the statement that we have proved is true or not, one can put the values of all the variables and check if DC = $\dfrac{h\left( \tan \alpha +\tan \beta \right)}{\tan \alpha \tan \beta }$ is true or not. Students must be able to understand the meaning of the question to solve it correctly.
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