# Question #2f5ac

Jun 13, 2016

(ii) $2.0 \times {10}^{4} \tan \alpha \text{ meter}$
(iii) $\sqrt{7} \times {10}^{4} \text{ meter}$
(iv) ${32}^{\circ}$, correct to nearest degree.

#### Explanation:

(i) My diagram looks like as below

(ii) In $\Delta A D C$, after converting all distances to meters.
$\tan \alpha = \frac{C D}{20000}$
$\implies C D = 2.0 \times {10}^{4} \tan \alpha \text{ meter}$
(iii) We make use of cosine rule.
For a triangle with sides $a , b \mathmr{and} c$ and angles $A , B \mathmr{and} C$ the cosine rule states:
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$
Applying in $\Delta A B C$ where $\angle A = {120}^{\circ}$
$B {C}^{2} = {\left(2.0 \times {10}^{4}\right)}^{2} + {\left(1.0 \times {10}^{4}\right)}^{2} - 2 \times 2.0 \times {10}^{4} \times 1.0 \times {10}^{4} \cos {120}^{\circ}$
We know that $\cos {120}^{\circ} = - \frac{1}{2}$

$\implies B {C}^{2} = 4.0 \times {10}^{8} + 1.0 \times {10}^{8} + 2.0 \times {10}^{8}$
$\implies B {C}^{2} = 7.0 \times {10}^{8}$
$\implies B C = \sqrt{7.0 \times {10}^{8}}$
$\implies B C = \sqrt{7} \times {10}^{4} \text{ meter}$
(iv) In $\Delta B D C$
$\tan {25}^{\circ} 15 ' = \frac{C D}{B C}$
Making use of values found in (ii) and (iii)
$\tan {25}^{\circ} 15 ' = \frac{2.0 \times {10}^{4} \tan \alpha}{\sqrt{7} \times {10}^{4}}$
Solving for $\alpha$
$\tan \alpha = \frac{\sqrt{7}}{2} \tan {25}^{\circ} 15 '$
$\alpha = {\tan}^{-} 1 \left(0.6239\right)$
$\alpha \approx {32}^{\circ}$, correct to nearest degree.