Ch $4$ . $4$ Solving Obtuse Triangle Situations $\newline$ Example $2$ : A rescue helicopter flew from its home base for $35 \mathrm{~km}$ on a bearing $\mathrm{S} 60^{\circ} \mathrm{E}$ to pick up an accident victim. Then it flew directly south for $25 \mathrm{~km}$ to the hospital. What distance and course (bearing from the hospital) must the helicopter fly to return directly to its home base? Answer to the nearest tenth of a kilometre and tenth of a degree. $\newline$ $\left(52.2 \mathrm{~km}, \mathrm{~N} 35.5^{\circ} \mathrm{W}\right)$

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Grade 6

Division with decimal divisors

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Q. Ch

4

4

Solving Obtuse Triangle Situations

\newline

Example

2

: A rescue helicopter flew from its home base for

35 \mathrm{~km}

on a bearing

\mathrm{S} 60^{\circ} \mathrm{E}

to pick up an accident victim. Then it flew directly south for

25 \mathrm{~km}

to the hospital. What distance and course (bearing from the hospital) must the helicopter fly to return directly to its home base? Answer to the nearest tenth of a kilometre and tenth of a degree.

\newline

\left(52.2 \mathrm{~km}, \mathrm{~N} 35.5^{\circ} \mathrm{W}\right)

Draw Situation: First, let's draw the situation to visualize the problem. We have a right-angled triangle where the helicopter first flew $35\,\text{km}$ on a bearing of $\text{S}60^\circ\text{E}$ and then flew directly south for $25\,\text{km}$ .
Calculate Hypotenize: To find the distance back to the base, we need to calculate the hypotenuse of the right-angled triangle. The sides we know are $35\,\text{km}$ (the adjacent side to the angle at the base) and $25\,\text{km}$ (the opposite side).
Use Pythagorean Theorem: We use the Pythagorean theorem to find the hypotenuse (distance back to base): $\text{hypotenuse}^2 = \text{adjacent}^2 + \text{opposite}^2$ .
Plug in Numbers: Plugging in the numbers: $\text{hypotenuse}^2 = 35^2 + 25^2 = 1225 + 625 = 1850$ .
Find Hypotenuse: Now, take the square root of $1850$ to find the hypotenuse: $\text{hypotenuse} = \sqrt{1850} \approx 43.0$ km.