Question

Solve the problem.A plane flying a straight course observes a mountain at a bearing of 35° to the right of its course. At that time the plane is 10 kilometres from the mountain. A short time later, the bearing to the mountain becomes 45°. How far is the plane from the mountain when the second bearing is taken (to the nearest tenth of a km)?Group of answer choices8.1 kilometers13.6 kilometers5.4 kilometers12.3 kilometers

Answer 1

Given:

A plane flying a straight course observes a mountain at a bearing of 35° to the right of its course.

The distance between plan and mountain is 10 km.

A short time later, the bearing to the mountain becomes 45°.

Here NM is the distance between the plane from the mountain when the second bearing is taken.

We need to find the measure of NM.

$We\text{ know that }\angle INM\text{ and }\angle\text{XNM are supplementary angles.}$

The sum of the supplementary angles is 180 degrees.

$\angle INM+\angle XNM=180^o$$\text{Substitute }\angle XNM=45^o\text{ in the equation.}$

$\angle INM+45^o=180^o$

$\angle INM=180^o-45^o$

$\angle INM=135^o$

We know that the sum of all three angles of the triangle is 180 degrees.

$\angle INM+\angle NIM+\angle INM=180^o$$\text{Substitute }\angle INM=135^o\text{ and }\angle NIM=35^o\text{ in the equation.}$

$135^o+35^o+\angle INM=180^o$

$170^o+\angle INM=180^o$

$\angle INM=180^o-170^o$

$\angle INM=10^o$

Consider the sine law.

$\frac{\sin N}{IM}=\frac{\sin M}{IN}=\frac{\sin I}{NM}$

Take the equation to find the measure of NM.

$\frac{\sin N}{IM}=\frac{\sin I}{NM}$$\text{Substitute }\angle N=135^o,\angle I=35^o,\text{ and IM=10 in the equation.}$

$\frac{\sin 135^o}{10}=\frac{\sin35^o}{NM}$

$NM=\frac{\sin 35^o}{\sin 135^o}\times10$$NM=8.11$

Hence the measure of NM is 8.1 km.

The plane is 8.1 km far from the mountain when the second bearing is taken.