Use your favorite paint program to highlight triangle ABD.
This is a right triangle because of the square angle marker at angle ADB.
The reference angle is 8 degrees.
Opposite the reference angle is AD = 10.
Adjacent to the reference angle is BD.
Let's use the tangent trig ratio to find side BD.
Focus on triangle ABD only.
tan(angle) = opposite/adjacent
tan(B) = AD/BD
tan(8) = 10/BD
BD*tan(8) = 10
BD = 10/tan(8)
BD = 71.153697223842
BD = 71.153697
This value is approximate.
Please make sure that your calculator is set to degrees mode.
Now let's highlight triangle ADC.
We'll use tangent once again since triangle ADC is a right triangle.
tan(angle) = opposite/adjacent
tan(C) = AD/CD
CD = AD/tan(C)
CD = 10/tan(11)
CD = 51.445540159703
CD = 51.445540
The summary so far is that we found these approximate side lengths
BD = 71.153697
CD = 51.445540
They will be used in the next part to find side BC.
The last thing to highlight is triangle BCD.
Use the Law of Cosines.
(BC)^2 = (BD)^2 + (CD)^2 - 2*(BD)*(CD)*cos(angle BDC)
(BC)^2 = (71.153697)^2 + (51.445540)^2 - 2*(71.153697)*(51.445540)*cos(85)
(BC)^2 = 7071.417954
BC = sqrt(7071.417954)
BC = 84.09172
BC = 84
I rounded the final answer to the nearest whole number since the 10 cm given to us is a whole number.
However, round that approximate value to some other level of precision if your teacher requires it.
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