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

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 PIm h class=

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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°