Question

section 2.5 - further applications of right triangles
4.) a plane flies 1.3 hours at 110 mph on a bearing of 38 degrees. it then turns and flies 1.5 hours at the same speed on a bearing of 128 degrees. how far is the plane from its starting point? see given

Explanation:

Step1: Calculate the first - part distance

The distance formula is $d = vt$. For the first part of the flight, $v = 110$ mph and $t=1.3$ hours. So, $d_1=110\times1.3 = 143$ miles.

Step2: Calculate the second - part distance

For the second part of the flight, $v = 110$ mph and $t = 1.5$ hours. So, $d_2=110\times1.5=165$ miles.

Step3: Find the angle between the two paths

The angle between the two bearings $\theta=128 - 38=90^{\circ}$.

Step4: Use the Pythagorean theorem

Since the two paths form a right - triangle, the distance $D$ from the starting point is given by $D=\sqrt{d_1^{2}+d_2^{2}}$. Substitute $d_1 = 143$ and $d_2 = 165$ into the formula: $D=\sqrt{143^{2}+165^{2}}=\sqrt{20449 + 27225}=\sqrt{47674}\approx218.35$ miles.

Answer:

Approximately 218.35 miles