Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

1. P, Q and R are three buildings. A car began its journey at P, drove to Q, then to R and returned to P. The bearing of Q from P is 058º and R is due east of Q. PQ = 114 km and QR = 70 km. © Draw a clearly labelled diagram to represent the above informationen on the diagram TƏRund (a) the north/south direction (b) the bearing 058° (c) the distances 114 km and 70 km. (ii) Calculate (a) the measure of angle POR (b) the distance PR [3] (c) the bearing of P from R [3]

1 P Q And R Are Three Buildings A Car Began Its Journey At P Drove To Q Then To R And Returned To P The Bearing Of Q From P Is 058º And R Is Due East Of Q PQ 11 class=

Sagot :

Step 1

Given;

[tex]\begin{gathered} The\text{ bearing of Q from P is 058}^o\text{ } \\ R\text{ is due east of Q} \\ PQ=114km \\ QR=70km \end{gathered}[/tex]

Step 2

Draw the diagram

Step 2

Calculate the measure of angle PQR

[tex]\angle PQR=58+90=148^o[/tex]

This is because using alternate exterior angles are equal theorem, the first part of angle Q 58 degrees. Since R is due east of Q, then the other part must be 90 degrees, when summed we get 148 degrees

Step 3

Calculate the distance PR. To do this we will use the cosine rule

[tex]\begin{gathered} PR^2=PQ^2+QR^2-2PQ\left(QR\right?cosQ \\ PR^2=114^2+70^2-2\left(114\right)\left(70\right)cos\left(148\right) \\ PR^2=17896+13534.84761 \\ PR=\sqrt{31430.84761} \\ PR=177.2874717 \\ PR\approx177.3km\text{ to the nearest tenth} \end{gathered}[/tex]

Step 4

Calculate the bearing of P from R.

Use sine rule and find angle R

[tex]\frac{sin\text{ 148}}{177.2874717}=\frac{sinR}{114}[/tex][tex]\begin{gathered} 114sin148=177.2874717sinR \\ R=\sin^{-1}\frac{\mleft(114sin148\mright)}{177.2874717} \\ R=19.92260569 \end{gathered}[/tex]

The bearing of P from R = (90-angle R)+90+90=250 degrees approximately to the nearest whole number

[tex]\begin{gathered} =\left(90-19.92260569\right)+90+90 \\ =250.07739 \\ \approx250^o \end{gathered}[/tex]

The bearing of P from R =250 degrees approximately to the nearest whole number

View image MaguireJ195653
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.