The diagram shows a tetrahedron. AD is perpendicular to both AB and AC.

AB = 10 cm.

AC = 8 cm.

AD = 5 cm.

Angle BAC = 90°

Calculate the size of angle BDC.

Give your answer correct to 1 decimal place.

The size of the angle ∠BDC of the tetrahedron can be determined by

cosine rule.

The size of angle ∠BDC is approximately 71.1°

Reasons:

A tetrahedron is a pyramid with triangular faces.

By Pythagoras's theorem, we have;

² = ² + ²

∴ ² = 8² + 5² = 89

² = 89

² = ² + ²

∴ ² = 10² + 8² = 164

² = 164

² = ² + ²

² = 5² + 10² = 150

² = 150

By cosine rule, we have;

² = ² + ² - 2 ×   × × cos(∠BDC)

Therefore;

164 = 150 + 89 - 2 × √(150) × √(89) × cos(∠BDC)

2 × √(150) × √(89) × cos(∠BDC) = 150 + 89 - 164 = 75

∠BDC ≈ 71.1°

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Work Shown:

For now focus on triangle ABC. Since this is a right triangle, we can use the pythagorean theorem to find the hypotenuse BC.

a^2 + b^2 = c^2

(AC)^2 + (AB)^2 = (BC)^2

8^2 + 10^2 = (BC)^2

(BC)^2 = 164

BC = sqrt(164)

BC = 12.806248 approximately

Move onto triangle ABD. Find the missing side BD through the pythagorean theorem

a^2 + b^2 = c^2

(AD)^2 + (AB)^2 = (BD)^2

5^2 + 10^2 = (BD)^2

125 = (BD)^2

(BD)^2 = 125

BD = sqrt(125)

BD = 11.1803399 which is approximate

Move onto triangle ADC. We'll use the pythagorean theorem again

a^2 + b^2 = c^2

(AC)^2 + (AD)^2 = (CD)^2

8^2 + 5^2 = (CD)^2

89 = (CD)^2

(CD)^2 = 89

CD = sqrt(89)

CD = 9.433981 also approximate

Focus on triangle BCD. We have the following side lengths

Let

Note the use of uppercase letters for angles and lowercase letters for side lengths.

We can then use the law of cosines to find the angle we're after

d^2 = b^2 + c^2 - 2*b*c*cos(D)

12.806248^2=9.433981^2+11.1803399^2-2*9.433981*11.1803399*cos(D)

163.999987837504=213.999997787893-210.950228380284*cos(D)

-210.950228380284*cos(D) = -50.000009950389

cos(D) = (-50.000009950389)/(-210.950228380284)

cos(D) = 0.237022781792173

cos(D) = arccos(0.237022781792173)

D = 76.2891111084541

D = 76.3

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