Question

The diagram shows a tetrahedron.

Answer 1

Answer: Angle BDC = 76.3 degrees

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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

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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

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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

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Focus on triangle BCD. We have the following side lengths

• BC = 12.806248
• BD = 11.1803399
• CD = 9.433981

Let

• side d be opposite angle D, so d = BC = 12.806248
• side b be opposite angle B, so b = CD = 9.433981
• side c be opposite angle C, so c = BD = 11.1803399

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