=================================================
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
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;
[tex]\overline{DC}[/tex]² = [tex]\mathbf{\overline{AD}}[/tex]² + [tex]\mathbf{\overline{AC}}[/tex]²
∴ [tex]\overline{DC}[/tex]² = 8² + 5² = 89
[tex]\overline{DC}[/tex]² = 89
[tex]\overline{BC}[/tex]² = [tex]\overline{AB}[/tex]² + [tex]\overline{AC}[/tex]²
∴ [tex]\overline{BC}[/tex]² = 10² + 8² = 164
[tex]\overline{BC}[/tex]² = 164
[tex]\overline{DB}[/tex]² = [tex]\mathbf{\overline{AD}}[/tex]² + [tex]\mathbf{\overline{AB}}[/tex]²
[tex]\overline{DB}[/tex]² = 5² + 10² = 150
[tex]\overline{DB}[/tex]² = 150
By cosine rule, we have;
[tex]\overline{BC}[/tex]² = [tex]\overline{DB}[/tex]² + [tex]\overline{DC}[/tex]² - 2 × [tex]\mathbf{\overline{DB}}[/tex] × [tex]\mathbf{\overline{DC}}[/tex] × cos(∠BDC)
Therefore;
164 = 150 + 89 - 2 × √(150) × √(89) × cos(∠BDC)
2 × √(150) × √(89) × cos(∠BDC) = 150 + 89 - 164 = 75
[tex]cos( \angle BDC) = \mathbf{\dfrac{75}{2 \times \sqrt{150} \times \sqrt{89} }}[/tex]
[tex]\angle BDC = arccos \left(\dfrac{75}{2 \times \sqrt{150} \times \sqrt{89} } \right) \approx \mathbf{71.1 ^{\circ}}[/tex]
∠BDC ≈ 71.1°
Learn more here:
https://brainly.com/question/20839703
