[Math] Calculating angle in a cuboid

trigonometry

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."

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I have used the cosine rule for this question.

Here is my working:

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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.

Best Answer

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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