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I have the question "Calculate the angle between the line BH and the plane ABCD in the cuboid pictured below, giving your answer to 1 decimal place."

enter image description here

I have used the cosine rule for this question.

Here is my working:

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enter image description here

So I created a triangle DBH and then found the sides BD and BH using Pythagoras.

I then put these values into the cosine rule and got the final answer of 84.6 degrees.

However the solutions say that the answer should be 34.5 degrees.

Have I used the wrong method ? Should I have used the sine rule ? I have not used the sine rule as there are no angles given.

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I get $2 \cdot \sqrt{53} \cdot \sqrt{78}\approx 128.59$ in the denominator. It looks like you did not take the square root of $78$.

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  • $\begingroup$ Yes that's correct. Thank you :) $\endgroup$ – Dan Mar 6 '17 at 17:09
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I don't understand why the cosines law: you have here cuboid = a right three dimensional box, so $\;\angle BDH=90^\circ\;$ and you can directly use trigonometry here:

$$\cos\theta=\frac{BD}{BH}=\sqrt\frac{53}{78}\approx82.43\implies \theta=\arccos(82.43)=34.48^\circ\approx34.5^\circ$$

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  • $\begingroup$ Yes I forgot to take the square root of 78 but thank you :) $\endgroup$ – Dan Mar 6 '17 at 17:11
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You are doing more work than necessary. Once you have the lengths of the sides, just use $\cos(\theta) = \frac{adjacent}{hypotenuse} = \frac{\sqrt{53}}{\sqrt{78}}$. Thus

$$\theta = \arccos\left(\frac{\sqrt{53}}{\sqrt{78}}\right) \approx 34.48^\circ$$

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  • $\begingroup$ Oh wow I didn't know you could do that thanks :) $\endgroup$ – Dan Mar 6 '17 at 17:11
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@Alexis Olson I think that a majority of people, when they have the choice, will opt for $\tan^{-1}=atan$ instead of $\cos^{-1}$, giving:

$\tan\theta=\frac{HD}{BD}=\frac{5}{\sqrt{53}}\approx 0.6868\implies \theta=atan(0.6868)=0.6018 rad = 34.48^\circ.$

The reason of this preference is that $atan$ has a "smoother behavior" than $acos=cos^{-1}.$

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