Aship sails from harbour h on a bearing of 084 for 340 km until it reaches point p. it the sails on a bearing of 210 for 160km until it reaches point q. calculate distance between point q and harbor. ( me find angle p, i do not know if i need to ise angle 210 to solve for qh)

QH = 227.8 km ≅ 228 km

Step-by-step explanation:

∵ The bearing from H to P is 084°

∵ The bearing from P to Q is 210°

∵ The distance from H to P = 340 km

∵ The distance from P to Q = 160 km

∴ The angle between 340 and 160 = 360 - 210 - (180 - 84) = 54°

( 180 - 84) ⇒ interior supplementary

By using cos Rule:

(QH)² = (PH)² + (PQ)² - 2(PH)(PQ)cos∠HPQ

(QH)² = 340² + 160² - 2(340)(160)cos(54) = 51904.965

∴ QH = 227.8 km ≅ 228 km

Angle at p is 54°

Distance between Q and harbor is

77249 km

Step-by-step explanation:

The bearing from H to P is 084°

∵ The bearing from P to Q is 210°

∵ The distance from H to P = 340 km

∵ The distance from P to Q = 160 km

∴ The angle at p = 84 - (210-180)

= 84 - 30

= 54°

Using cosine rule

And let's call length qh "x"

X = (hp)² + (pq)² -2(hp)(pq)cosx

X = 340²+160² -2(340)(160)cos54

X = 115600+25600 -2(54400)(0.587785)

X= 141200-63951

X = 77249km

7

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car

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