1. **Problem statement:** We have a triangle PQR with sides PQ = 5 m, QR = 13 m, and angle PÔQ = 90°. Q is on a bearing of 126° from P. 2. **(a)(i) Find the bearing of P from Q:** - Bearing from P to Q is 126°. - Bearing from Q to P is the opposite direction, so add 180°: $$126^\circ + 180^\circ = 306^\circ$$ 3. **(a)(ii) Find the bearing of P from R:** - Given bearing from P to Q is 126°. - The bearing of P from R is calculated as: $$180^\circ - 126^\circ = 54^\circ$$ 4. **(b)(i) Work out length PR:** - Using the cosine rule in triangle PQR: $$PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PÔQ)$$ - Since $\angle PÔQ = 90^\circ$, $\cos 90^\circ = 0$, so: $$PR^2 = 5^2 + 13^2 = 25 + 169 = 194$$ - Therefore: $$PR = \sqrt{194} \approx 13.928$$ 5. **(b)(ii) Write down the value of $\cos PÔR$ as a fraction:** - The angle $PÔR$ corresponds to the difference in bearings between P and R from O. - Given bearings 306° and 360°, the fraction is: $$\frac{306}{360} = \frac{17}{20}$$ **Final answers:** - (a)(i) Bearing of P from Q = $306^\circ$ - (a)(ii) Bearing of P from R = $54^\circ$ - (b)(i) Length PR $\approx 13.928$ m - (b)(ii) $\cos PÔR = \frac{17}{20}$