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ABCDEF GH is a cuboid - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 2

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ABCDEF GH is a cuboid. AB = 7.3 cm CH = 8.1 cm Angle BCI = 48° Find the size of the angle between AH and the plane ABCD. Give your answer correct to 1 decimal plac... show full transcript

Worked Solution & Example Answer:ABCDEF GH is a cuboid - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 2

Step 1

Find AC

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Answer

To find the length of AC, we can use the Pythagorean theorem. In triangle ABC:

AC=AB2+BC2AC = \sqrt{AB^2 + BC^2}

We know that:

  • AB=7.3 cmAB = 7.3 \text{ cm}
  • BC=CH=8.1 cmBC = CH = 8.1 \text{ cm}

Thus,

AC=(7.3)2+(8.1)2=53.29+65.61=118.9≈10.9 cmAC = \sqrt{(7.3)^2 + (8.1)^2} = \sqrt{53.29 + 65.61} = \sqrt{118.9} \approx 10.9 \text{ cm}

Step 2

Calculate angle CAH

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Answer

To find the angle CAH, we can use the cosine rule:

cos⁡(CAH)=BCAC=CHAH\cos(CAH) = \frac{BC}{AC} = \frac{CH}{AH}

From the previous calculations, we have:

  • AC≈10.9 cmAC \approx 10.9 \text{ cm}
  • CH=8.1 cmCH = 8.1 \text{ cm}

Now applying the cosine formula:

cos⁡(CAH)=8.110.9\cos(CAH) = \frac{8.1}{10.9}

To find the angle:

CAH=cos⁡−1(8.110.9)CAH = \cos^{-1}\left(\frac{8.1}{10.9}\right)

Step 3

Final calculation for angle between AH and plane ABCD

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Answer

To find the angle between line AH and the plane ABCD, we use:

Angle=90°−CAH\text{Angle} = 90° - CAH

Once CAH is calculated from the previous step, substitute the value to get:

Angle=90°−CAH\text{Angle} = 90° - CAH

Therefore, if CAH is calculated to be approximately X°, then:

Angle≈90°−X°\text{Angle} \approx 90° - X°

This should yield the final angle to one decimal place.

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