Solve Applied Problems Using the Law of Cosines
-A plane flying a straight course observes a mountain at a bearing of $31.7 ^ { \circ }$ to the right of its course. At that time the plane is 8 kilometers from the mountain. A short time later, the bearing to the mountain becomes $41.7 ^ { \circ }$. How far is the plane from the mountain when the second bearing is taken (to the nearest tenth of a $\mathrm { km } ) ?$
A) $6.3$ kilometers
B) $10.1$ kilometers
C) $11.4$ kilometers
D) $4.2$ kilometers

Question

Quizplus · Accepted Answer

The Law of Cosines states that in a triangle with sides $a$, $b$, and $c$, and opposite angles $A$, $B$, and $C$, respectively,$$c^2 = a^2 + b^2 - 2ab \cos C$$In this problem, we have a triangle with sides $a$ (the distance from the plane to the mountain when the first bearing is taken), $b$ (the distance from the plane to the mountain when the second bearing is taken), and $c$ (the distance the plane travels between the two bearings). We also know the angles $A$ (the bearing of the mountain when the first bearing is taken), $B$ (the bearing of the mountain when the second bearing is taken), and $C$ (the angle between the two bearings).We can use the Law of Cosines to solve for $b$:$$b^2 = a^2 + c^2 - 2ac \cos B$$We know that $a = 8$ kilometers, $c$ is the distance the plane travels between the two bearings, and $B = 41.7^\circ - 31.7^\circ = 10^\circ$. We can find $c$ by using the formula for the distance traveled by a plane:$$c = rt$$where $r$ is the speed of the plane and $t$ is the time it takes to travel between the two bearings. We don't know the speed of the plane, but we can assume that it is constant. Therefore, we can write:$$c = k \cdot t$$where $k$ is a constant.Substituting this into the Law of Cosines, we get:$$b^2 = a^2 + (k \cdot t)^2 - 2a(k \cdot t) \cos B$$We can simplify this to:$$b^2 = a^2 + k^2 t^2 - 2ak \cdot t \cos B$$We don't know the values of $k$ or $t$, but we can eliminate them by using the fact that the plane is flying a straight course. This means that the angle between the two bearings is equal to the angle between the two sides of the triangle that are adjacent to the angle $C$. Therefore, we have:$$\angle C = 180^\circ - (A + B) = 180^\circ - (31.7^\circ + 10^\circ) = 138.3^\circ$$We can now use the Law of Cosines to solve for $b$:$$b^2 = a^2 + c^2 - 2ac \cos C$$$$b^2 = 8^2 + (k \cdot t)^2 - 2(8)(k \cdot t) \cos 138.3^\circ$$$$b^2 = 64 + k^2 t^2 + 16k \cdot t \cos 138.3^\circ$$We don't know the values of $k$ or $t$, but we can find the value of $b$ by using the fact that the plane is flying a straight course. This means that the distance between the two bearings is equal to the distance the plane travels between the two bearings. Therefore, we have:$$b = c = k \cdot t$$Substituting this into the equation for $b^2$, we get:$$b^2 = 64 + k^2 t^2 + 16k \cdot t \cos 138.3^\circ$$$$b^2 = 64 + b^2 + 16b \cos 138.3^\circ$$$$b^2 - 16b \cos 138.3^\circ = 64$$$$b^2 - (-12.8b) = 64$$$$b^2 + 12.8b = 64$$$$(b + 6.4)^2 = 64 + 6.4^2 = 85.76$$$$b + 6.4 = \sqrt{85.76} = 9.27$$$$b = 9.27 - 6.4 = 2.87$$Therefore, the plane is $2.87$ kilometers from the mountain when the second bearing is taken.

Solve Applied Problems Using the Law of Cosines -A Plane 31.7∘31.7 ^ { \circ }31.7∘

Solve Applied Problems Using the Law of Cosines
-A Plane $31.7 ^ { \circ }$