QUESTION IMAGE
Question
2 résolvez chacun des triangles suivants.
a)
b
16 cm 94°
a 52° c
b)
e
22 cm 24 cm
d 62° f
Step1: Calculer l'angle \( C \)
Dans le triangle \( ABC \), la somme des angles d'un triangle est \( 180^{\circ} \).
On a \( \angle A = 52^{\circ} \), \( \angle B=94^{\circ} \).
Alors \( \angle C=180^{\circ}-\angle A - \angle B=180^{\circ}-52^{\circ}-94^{\circ}=34^{\circ} \)
Step2: Appliquer la loi des sinus
La loi des sinus est \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \)
On a \( c = AB = 16\mathrm{cm} \), \( \angle A = 52^{\circ} \), \( \angle B = 94^{\circ} \), \( \angle C=34^{\circ} \)
Pour trouver \( a = BC \):
\( \frac{a}{\sin A}=\frac{c}{\sin C}\)
\( a=\frac{c\times\sin A}{\sin C}=\frac{16\times\sin52^{\circ}}{\sin34^{\circ}}\)
\( \sin52^{\circ}\approx0.7880 \), \( \sin34^{\circ}\approx0.5592 \)
\( a=\frac{16\times0.7880}{0.5592}\approx22.5\mathrm{cm} \)
Pour trouver \( b = AC \):
\( \frac{b}{\sin B}=\frac{c}{\sin C}\)
\( b=\frac{c\times\sin B}{\sin C}=\frac{16\times\sin94^{\circ}}{\sin34^{\circ}}\)
\( \sin94^{\circ}\approx0.9976 \)
\( b=\frac{16\times0.9976}{0.5592}\approx28.5\mathrm{cm} \)
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Dans le triangle \( ABC \):
- \( \angle C = 34^{\circ} \)
- \( BC\approx22.5\mathrm{cm} \)
- \( AC\approx28.5\mathrm{cm} \)