tom.rizo

New Member
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

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MarkFL

La Villa Strangiato
Staff member
If we apply a bit of trigonometry here, and we let $$h$$ be the altitude we seek, we may state:
$$\displaystyle \sin\left(42^{\circ}\right)=\frac{h}{10}\implies h=10\sin\left(42^{\circ}\right)\approx6.7$$