Math Help Forum - View Single Post

magentarita · #5 January 6th, 2009, 10:57 PM

Quote:

Originally Posted by earboth

1. $45' = \left(\dfrac{45}{60}\right)^\circ = 0.75^\circ$

2. You are dealing with 2 right triangles. Use tan-function to calculate the second leg:

The length of the flagpole is:

$f = 100\cdot \tan(39.75^\circ) - 100\cdot \tan(28^\circ) = 100 \cdot (\tan(39.75^\circ) - \tan(28^\circ)) = 29.998...\approx 30'$

Nice picture and reply.