Answer:
13.6 miles (nearest tenth)
Step-by-step explanation:
The given scenario can be modelled as triangle ABC, where the location of the glider plane is at vertex A, and the two radar stations are located at vertices B and C.
Given that the two radar stations are 30 miles apart, BC = 30.
If the angle of elevation from the radar station to the left of the plane is 25°, and the angle of elevation from the radar station to the right of the plane is 18°, then m∠B = 25° and m∠C = 18°.
As the interior angles of a triangle sum to 180° then:
m∠A + m∠B + m∠C= 180°
m∠A + 25° + 18° = 180°
m∠A + 43° = 180°
m∠A = 137°
To find the distance between the left radar station and the plane, we need to calculate the length of side AB, which is the side opposite angle C. To do this, we can use the Law of Sines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
In this case:
- A = 137°
- B = 25°
- C = 18°
- a = 30
- b = AC
- c = AB
Substitute the values into the equation:
[tex]\dfrac{30}{\sin 137^{\circ}}=\dfrac{AC}{\sin 25^{\circ}}=\dfrac{AB}{\sin 18^{\circ}}[/tex]
As we want to solve for AB, we can discard the middle fraction to give us:
[tex]\dfrac{30}{\sin 137^{\circ}}=\dfrac{AB}{\sin 18^{\circ}}[/tex]
Now, solve for AB:
[tex]\dfrac{AB}{\sin 18^{\circ}}=\dfrac{30}{\sin 137^{\circ}}\\\\\\\\AB=\dfrac{30\sin 18^{\circ}}{\sin 137^{\circ}}\\\\\\AB=13.5931556058...\\\\\\AB=13.6\; \sf miles\;(nearest\;tenth)[/tex]
Therefore, the distance between the left radar station and plane is:
[tex]\LARGE\boxed{\boxed{13.6\; \sf miles}}[/tex]
