7. In the figure, P, Q and R points on level ground in a park. R is 600 m from Q and on a bearing of 305° from Q. Q is 950 m from P and on a bearing of 078° from P
Im having trouble with finding the distance. a
Answers
Answer:
The answer is below
Step-by-step explanation:
From the image:
∠P + ∠Q = 180° (consecutive angles are supplementary)
78 + ∠Q = 180
∠Q = 180 - 78
∠Q = 102°
∠RQN = 360° - 305° =(sum of angle in a point)
∠RQN = 55°
∠RQN + ∠RQP = 102°
∠RQP = 102 - 55 = 47°
a) Let PR = x
Using cosine rule:
[tex]x^2=600^2+950^2-2(900)(600)cos(47)\\\\x^2=485021.9\\\\x=\sqrt{485021.9} \\\\x=696 \ m[/tex]
Therefore PR = 696 m
b) Let ∠RPQ be A°
Using sine rule:
[tex]\frac{sin(A)}{600} =\frac{sin(47)}{696} \\\\sin(A)=\frac{sin(47)}{696} *600\\\\sin(A)=0.6304\\\\A=sin^{-1}(0.6304)\\\\A=39^0[/tex]
∠RPQ = 39°
Hence the bearing of R from P = ∠NPR = 78° - 39° = 39°
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