Question

In the above figure, point $C$ is approximately $18$ meters from $B$. The measure of $\angle CBA$ is ${45}^o$, and the measure of $\angle BCA$ is ${60}^o$.
Which of the following expressions gives the approximate distance in meters between point $A$ and point $B$

• A. $\frac{\sin {60}^o}{18\sin {45}^o}$
• B. $\frac{\sin {60}^o}{18\sin {75}^o}$
• C. $\frac{18\sin {45}^o}{\sin {60}^o}$
• D. $\frac{18\sin {60}^o}{\sin {45}^o}$
• E. $\frac{18\sin {60}^o}{\sin {75}^o}$

Answer 1

The correct option is E $\frac{18\sin {60}^o}{\sin {75}^o}$

Applying Sine rule gives us
$\frac{BC}{\sin A}=\frac{AB}{\sin C}=\frac{AC}{\sin B}$
Hence, $AB=\frac{BC\left(\sin C\right)}{\sin A}$

$=\frac{18\sin C}{\sin A}$

$=\frac{18\sin {60}^0}{\sin ({180}^0-(B+C\left)\right)}$

$=\frac{18\sin {60}^0}{\sin ({180}^0-{105}^0)}$

$=\frac{18\sin {60}^0}{\sin {75}^0}$