<h3>
Answer: Angle BDC = 76.3 degrees</h3>
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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
