How do you solve the triangle given $△ D E F, d = 72, e = 48, m ∠ F = 110$ $triangleDEF, d=72, e=48, mangleF=110$ ?

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Answer 1 · 2018-02-26T08:36:50

$f = 99.2575, ˆ D = 43^{\circ}, ˆ E = 27^{\circ}$ $f = 99.2575, hat D = 43^@, hat E = 27^@$

Applying law of Cosines,

$f = \sqrt{d^{2} + e^{2} - 2 d e cos F}$ $f = sqrt(d^2 + e^2 - 2 d e cos F )$

$\Rightarrow \sqrt{72^{2} + 48^{2} - (2 \cdot 72 \cdot 48 \cdot cos 110)} = 99.2575$ $=> sqrt(72^2 + 48^2 - (2 * 72 * 48 * cos 110)) = 99.2575$

Applying law of sines,

$\frac{d}{sin D} = \frac{e}{sin E} = \frac{f}{sin F}$ $d / sin D = e / sin E = f / sin F$

$sin D = \frac{d \cdot sin F}{f} = \frac{72 \cdot sin 110}{99.2575} = 0.6816$ $sin D = (d * sin F) / f = (72 * sin 110) / 99.2575 = 0.6816$

$ˆ D = {sin}^{- 1} 0.6816 = 43^{\circ}$ $hatD = sin ^-1 0.6816 = 43^@$

$ˆ E = 180 - 110 - 43 = 27^{\circ}$ $hat E = 180 - 110 - 43 = 27^@$

How do you solve the triangle given △DEF,d=72,e=48,m∠F=110triangleDEF, d=72, e=48, mangleF=110?