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secondary 4 | E Maths
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I need help with part b, thank you
Date Posted: 3 years ago
Views: 562
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Since △PQR is isosceles, PQ = RQ
You've already found that ∠PQR = 30°
Area of △PQR = ½ ab sin c
= ½ (PQ)(RQ) sin ∠PQR
= ½ (16cm)(16cm) sin 30°
= (128cm²) (½)
= 64cm²
You should be able to do from here.
You've already found that ∠PQR = 30°
Area of △PQR = ½ ab sin c
= ½ (PQ)(RQ) sin ∠PQR
= ½ (16cm)(16cm) sin 30°
= (128cm²) (½)
= 64cm²
You should be able to do from here.
By the way, please alert your teacher to the following :
There is an error in the diagram. It is mathematically incorrect.
In an isosceles triangle with sides 16cm,16cm and 7cm, the angle PQR cannot be 30°. It should be ≈ 25.271° instead.
Otherwise, if the angle 30° is taken to be correct, the side PQ cannot equal 16cm. It should be ≈ 13.523 cm.
Solving the question with other methods (eg. Cosine Rule or Sine Rule) would result in errors if both 30° and 16cm are used, as sine and cosine of some angles would be greater than 1 or less than -1.
And we know that -1 ≤ sin θ ≤ 1and -1≤ sin θ ≤ 1 for all θ
There is an error in the diagram. It is mathematically incorrect.
In an isosceles triangle with sides 16cm,16cm and 7cm, the angle PQR cannot be 30°. It should be ≈ 25.271° instead.
Otherwise, if the angle 30° is taken to be correct, the side PQ cannot equal 16cm. It should be ≈ 13.523 cm.
Solving the question with other methods (eg. Cosine Rule or Sine Rule) would result in errors if both 30° and 16cm are used, as sine and cosine of some angles would be greater than 1 or less than -1.
And we know that -1 ≤ sin θ ≤ 1and -1≤ sin θ ≤ 1 for all θ
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