Answered

Discover a world of knowledge and get your questions answered at IDNLearn.com. Discover in-depth and trustworthy answers from our extensive network of knowledgeable professionals.

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
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.