a cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farm respectively. If Dogbe walked 5km and Manu 3km from the cottage to their farm, find and correct to 2 s.f the distance between the two farms and correct to the nearest degree, the bearing of Manu's farm from Dogbe's
PLEASE NOTE: DOGBE AND MANU ARE NAMES
Answers
Based on the drawing obtained from the description, the location of
Manu's farm is located relatively south to Dogbe's farm.
The distance between the two farms is approximately 3.8 kmThe bearing of Manu's farm from Dogbe is approximately 238°Reasons:
The bearing of the cottage to Dogbe's farm = 200°
The distance Dogbe walks from the cottage to his farm = 5 km
The bearing of the cottage from Manu's farm = 110°
The distance Manu walks from the cottage to his farm = 3 km
Required:
The distance and between the two farms.
Solution:
Please find attached a drawing showing the position of the cottage
By cosine rule, we have;
a² = b² + c² - 2·b·c·cos(A)Where;
b = The distance Dogbe walks from the cottage to his farm = 5 km
c = The distance Manu walks to his farm from the cottage = 3 km
a = The distance between the two farms = d
A = The angle between the paths to the cottage from the farms = 50°
By plugging in the values, we have;
d² = 5² + 3² - 2 × 5 × 3 × cos(50°)
d = √(5² + 3² - 2 × 5 × 3 × cos(50°)) ≈ 3.8
The distance between the two farms, d ≈ 3.8 kmRequired:
The bearing of Manu's farm from Dogbe's
Solution:
By sine rule, we have;
[tex]\displaystyle \frac{3}{sin(C)} = \mathbf{ \frac{d}{sin(50^{\circ})}}[/tex]Which gives;
[tex]\displaystyle sin(C) = \mathbf{\frac{3 \times sin(50^{\circ})}{\sqrt{5^2 + 3^2 - 2 \times 5 \times 3 \times cos(50^{\circ})} }}[/tex]
[tex]\displaystyle \angle C = arcsine \left(\frac{3 \times sin(50^{\circ})}{\sqrt{5^2 + 3^2 - 2 \times 5 \times 3 \times cos(50^{\circ})} } \right) \approx 38 ^{\circ}[/tex]
The bearing of Manu's farm from Dogbe's = 200° + ∠C
Therefore;
The bearing of Manu's farm from Dogbe's ≈ 200° + 38 = 238°Learn more about bearings in mathematics here:
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[tex]\huge\text{\bf Hey there!}[/tex]
[tex]\huge\text{GUIDE} \ \boxed{\downarrow}\\\large\boxed{\mathsf{>: is \ \bf greater \ than }}\\\\\large\boxed{\mathsf{<: is \bf \ less\ than}}\\\\\large\boxed{\mathsf{\geq: is \ \bf greater\ than\ or\ equal\ to}}\\\\\large\boxed{\mathsf{\leq: is\ \bf less\ than\ or\ equal\ to}}\\\\\large\boxed{\mathsf{=: is \ \bf equal \ to}}[/tex]
[tex]\huge\text{GUIDE \#2}\ \boxed{\downarrow}\\\large\text{When you are working with greater than, you're basically shows}\\\large\text{when a number is BIGGER than the other number.}\\\\\large\text{When you are working with less than you are finding the opposite}\\\large\text{of greater than. (it shows when a number is SMALLER than the}\\\large\text{number.)}[/tex]
[tex]\\\large\text{When you are working with greater than or equal to, you are }\\\large\text{basically finding out if the number is greater or equal to the other }\\\large\text{number.}\\\\\large\text{When you are working with less than or equal to you are trying to}\\\large\text{figure out if your number is less than or equal to the other}\\\large\text{number.}[/tex]
[tex]\huge\text{ANSWERING YOUR GIVEN QUESTION}\ \boxed{\downarrow}[/tex]
[tex]\large\boxed{\rm{-0.75 \ ? \ -\dfrac{3}{4}}}[/tex]
[tex]\large\text{I'm going to make this easier to solve for you. I will convert the}\\\large\text{fraction }(\rm{-\dfrac{3}{4}})\large\text{ into a decimal. To do so, you have to DIVIDE the}\\\large\text{numerator (the number on TOP) from the denominator (the}\\\large\text{BOTTOM number)}[/tex]
[tex]\large\text{REMEMBER }\textbf{!}\\\large\text{Negatives are BELOW zero. (including decimals \& fractions).}\\\\\large\text{Positives are ABOVE zero}[/tex]
[tex]\large\boxed{{\mathsf{- \dfrac{3}{4}}}}\\\\\\\large\boxed{\mathsf{\rightarrow - 3\div 4}}\\\\\\\large\boxed{\rightarrow\mathsf{-0.75}}[/tex]
[tex]\large\text{Therefore, -0.75 and }\rm{-\dfrac{3}{4}}\large\text{ are equivalent to each other}[/tex]
[tex]\huge\boxed{\text{Therefore, your answer is: }}\ \huge\boxed{\mathsf{-0.75 \bold{\ =} -\dfrac{3}{4}}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
Answer:
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Step-by-step explanation:
