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Determine the values of the unlabeled parts of the triangle.

OA) ZQ = 63.4°, PQ = 1.84, QO = 4.47

OB) ZQ = 26.6°, PQ = 2, QO = 5.12

OC) ZQ = 63.4º, PQ = 2, QO = 4.47

D)

ZQ = 63.4°, PQ = 1.84, QO = 5.12

User Jaison
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1 Answer

6 votes

Answer:

∠Q = 63.4º, PQ = 2, QO = 4.47

Explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

Triangle POQ is a right angled triangle because ∠P = 90°. Also, ∠O = 26.6°.

Therefore in ΔPOQ: ∠P + ∠O + ∠Q = 180° (sum of angles in a triangle).

90 + 26.6 + ∠Q = 180

∠Q + 116.6 = 180

∠Q = 180 - 116

∠Q = 63.4°

We can find PQ and QO using sine rule. Sine rule for ΔPOQ is given as:


(PO)/(sin Q)= (PQ)/(sinO)= (QO)/(sin P)\\\\Finding\ PQ :\\\\(PO)/(sin Q)= (PQ)/(sinO)\\\\(4)/(sin(63.4)) =(PQ)/(sin(26.6))\\\\PQ= (4)/(sin(63.4)) *sin(26.6)\\\\PQ = 2\\\\Also\ finding\ QO:\\\\(PO)/(sin Q)= (QO)/(sinP)\\\\(4)/(sin(63.4)) =(QO)/(sin(90))\\\\QO=(4)/(sin(63.4)) *sin(90)\\\\QO = 4.47

Determine the values of the unlabeled parts of the triangle. OA) ZQ = 63.4°, PQ = 1.84, QO-example-1
User David Tran
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