# Question 632ef

Apr 29, 2017

$\textcolor{red}{{34}^{o}}$

#### Explanation: We can construct this diagram from the given information.

Let length of $A B$ be ${l}_{1}$ and length of $B C$ be ${l}_{2}$.

It is given that ${l}_{1} / {l}_{2} = \frac{2}{3}$ $\to$ 1.

Also, observe that $B D = F E$ and $B F = D E$. $\to$ 2.

We have to find $\angle C A E$ which will give us elevation of C from A.

Let $\angle C A E$ be $\theta$.

From $\triangle A B F$ we observe that:-

$\frac{B F}{A B} = \sin {25}^{o} \implies B F = A B \cdot \sin {25}^{o} = 0.423 \cdot {l}_{1}$

$\therefore$ using 2., $D E = 0.423 \cdot {l}_{1}$ $\to$ 3.

$\frac{A F}{A B} = \cos {25}^{o} \implies A F = A B \cdot \cos {25}^{o} = 0.906 \cdot {l}_{1}$ $\to$ 4.

From $\triangle B C D$ we find:-

$\frac{C D}{B C} = \sin {40}^{o} \implies C D = B C \cdot \sin {40}^{o} = 0.643 \cdot {l}_{2}$ $\to$ 5.

$\frac{B D}{B C} = \cos {40}^{0} \implies B D = B C \cdot \cos {40}^{o} = 0.766 \cdot {l}_{2}$

$\therefore$ form 2., $F E = 0.766 \cdot {l}_{2}$ $\to$ 6.

Now,
$\tan \theta = \frac{C E}{A E} = \frac{C D + D E}{A F + F E}$

Substituting values of $D E , A F , C D , F E$ from 3., 4., 5., 6.

$\tan \theta = \frac{0.643 \cdot {l}_{2} + 0.423 \cdot {l}_{1}}{0.766 \cdot {l}_{2} + 0.906 \cdot {l}_{1}}$

Dividing the numerator and deniminator by ${l}_{2}$

$\implies \tan \theta = \frac{0.643 \cdot \cancel{{l}_{2} / {l}_{2}} + 0.423 \cdot {l}_{1} / {l}_{2}}{0.766 \cdot \cancel{{l}_{2} / {l}_{2}} + 0.906 \cdot {l}_{1} / {l}_{2}}$

Substituting value of ${l}_{1} / {l}_{2}$ from 1.

$\implies \tan \theta = \frac{0.643 + 0.423 \cdot \frac{2}{3}}{0.766 + 0.906 \cdot \frac{2}{3}} = 0.675$

$\therefore \theta = {\tan}^{-} 1 0.675 \approx {34}^{o}$

$\therefore \textcolor{red}{\angle C A E = {34}^{o}}$ which is the angle of elevation of $C$ from $A$.

Apr 29, 2017

The angle of elevation is: hatA = 34°

#### Explanation:

I am referring to the same image as posted by Veerpaksh S.
Thank you. The required angle of elevation of $C$ from $A$ can be seen as
$\hat{A}$ in $\Delta C A E$, a right-angled triangle.

The opposite side $\textcolor{b l u e}{C E = C D + D E}$

The adjacent side $\textcolor{red}{A E = A F + F E}$

You do not need the actual lengths of any of the lines, just a ratio is good enough. $A B : B C = 2 : 3$

In Delta CDB: " " (CD)/(BC) = sin 40°, :. color(blue)(CD = 3sin40°)

In Delta BFA: " "(BF)/(AB) = sin25°, :. color(blue)(BF = 2sin25°)

color(blue)(CE = 3sin40°+2sin25°)

In Delta CDB: " " (BD)/(BC) = cos 40°, :. color(red)(CD=3cos40°)

In Delta BFA: " "(AF)/(AB) = cos25°, :. color(red)(BF=2cos25°)

color(red)(AE = 3cos40°+2cos25°)

In $\Delta C A E :$

 tan A = (color(blue)(CE))/(color(red)(AE))= color(blue)((3sin40°+2sin25°))/(color(red)((3cos40°+2cos25°))

$\tan A = 0.6747$

hatA = 34°#