The average beam energy should be ((.3)(30)+(.6)(50)+(.1)(70)) = 46keV, because the fractions should add up to 1.0. However, the consequences are minor, and it will be OK if you use 1.9 cm$^{-1}$.

I realize that, given the limited time, B-5 could use one additional hint. While this is a pure geometry problem, there are multiple ways to approach it.

One path to the solution is to calculate two auxiliary points: The intersection of the LOC with the x-axis (call it maybe $x_i$) and the point on the LOC closest to the origin (call it maybe $(x_1, y_1)$). Next, you can easily determine the length of the triangle base from $(x_i, 0)$ to $(x_0, y_0)$ and derive the equation for $y_0$. The rest follows by substitution.

Minor mistake in Question D-1:

The average beam energy should be ((.3)(30)+(.6)(50)+(.1)(70)) = 46keV, because the fractions should add up to 1.0. However, the consequences are minor, and it will be OK if you use 1.9 cm$^{-1}$.

Thanks to David Liaguno for pointing this out.

I realize that, given the limited time, B-5 could use one additional hint. While this is a pure geometry problem, there are multiple ways to approach it.

One path to the solution is to calculate two auxiliary points: The intersection of the LOC with the x-axis (call it maybe $x_i$) and the point on the LOC closest to the origin (call it maybe $(x_1, y_1)$). Next, you can easily determine the length of the triangle base from $(x_i, 0)$ to $(x_0, y_0)$ and derive the equation for $y_0$. The rest follows by substitution.