T tom.rizo New Member Nov 19, 2018 1 0 1 Nov 19, 2018 #1 In acute triangle DEF, DE = 10 and angle D = 42 degrees. Find to the nearest tenth of the length of the altitude drawn to side DF. Please explain steps if possible, I want to understand how the solution was reached, thanks Last edited by a moderator: Nov 21, 2018

In acute triangle DEF, DE = 10 and angle D = 42 degrees. Find to the nearest tenth of the length of the altitude drawn to side DF. Please explain steps if possible, I want to understand how the solution was reached, thanks

MarkFL La Villa Strangiato Staff member Administrator Moderator Math Helper Jan 25, 2018 3,487 4,264 113 St. Augustine Nov 19, 2018 #2 If we apply a bit of trigonometry here, and we let \(h\) be the altitude we seek, we may state: \(\displaystyle \sin\left(42^{\circ}\right)=\frac{h}{10}\implies h=10\sin\left(42^{\circ}\right)\approx6.7\) Does this make sense? Reactions: TheJason and anemone

If we apply a bit of trigonometry here, and we let \(h\) be the altitude we seek, we may state: \(\displaystyle \sin\left(42^{\circ}\right)=\frac{h}{10}\implies h=10\sin\left(42^{\circ}\right)\approx6.7\) Does this make sense?