Dive into the world of knowledge and get your queries resolved at IDNLearn.com. Our platform is designed to provide accurate and comprehensive answers to any questions you may have.

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