Hey guys,
Can someone help me with this trig problem. The answer is 8.14 m cubed but I don't know how you work it out. Thanks!!!
I've redrawn the image (below) so I can refer to each vertex appropriately.
Length \(AD\) can be found using the cosine rule:
\left(125\right)\cos{\left(100^\circ\right)}\\AD&\approx164.22803617\text{ m}\end{align*})
We can then find \(\angle{ADB}\) using the sine or cosine rule:
\left(DB\right)\cos{\left(\angle{ADB}\right)}\\\cos{\left(\angle{ADB}\right)}&=-\frac{87^2-AD^2-125^2}{2\left(AD\right)\left(125\right)}\\\angle{ADB}&\approx31.446561128748^\circ\end{align*})
And we can therefore find \(\angle{ADC}\):
^\circ\\&\approx13.553438871255^\circ\end{align*})
Thus, we can now find \(\angle{CAD}\):
^\circ\\&=\left(180-111-\angle{ADC}\right)^\circ\\&\approx55.446561128745^\circ\end{align*})
Now we can find length \(CD\) using the sine rule:
}}&=\frac{AD}{\sin{\left(\angle{ACD}\right)}}\\\frac{CD}{\sin{\left(\angle{CAD}\right)}}&=\frac{AD}{\sin{\left(110^\circ\right)}}\\CD&=\frac{AD\sin{\left(\angle{CAD}\right)}}{\sin{\left(110^\circ\right)}}\\CD&\approx143.93835343452\text{ m}\end{align*})
Next, we can find the areas of triangles \(\Delta{ACD}\) and \(\Delta{ABD}\):
&=\frac{1}{2}\left(AD\right)\left(CD\right)\sin{\left(\angle{ADC}\right)}\\&\approx2769.8919315998\text{ m}^2\\\text{Area}\left(\Delta{ABD}\right)&=\frac{1}{2}\left(AB\right)\left(BD\right)\sin{\left(\angle{ABD}\right)}\\&=\frac{1}{2}\left(87\right)\left(125\right)\sin{\left(100^\circ\right)}\\&\approx5354.892157004\text{ m}^2\end{align*})
The total area is therefore \(\text{Area}\left(\Delta{ABD}\right)+\text{Area}\left(\Delta{ACD}\right)\) and the total volume is this area multiplied by 1 mm (which is \(\frac{1}{1000}\) m):
+\text{Area}\left(\Delta{ACD}\right)\right)\\&\approx8.1247840886038\text{ m}^3\end{align*})
Note a few things:
- The answer you have may be subject to some rounding errors
- It is for this reason that I have given my intermediate values to so many decimal places
- I stored these values in exact form (which were VERY ugly) in my CAS, rather than using the decimal approximations (for instance, \(\angle{CAD}=\left(\frac{180\sin^{-1}{\left(\frac{87\cos{\left(\frac{\pi}{18}\right)}\sqrt{2}}{2\sqrt{10875\sin{\left(\frac{\pi}{18}\right)}+11597}}\right)}}{\pi}+24\right)^\circ\) just as an example (also keep in mind, my calculator is in radians, rather than degrees)).
Home
Help
Search
Login
