Answer:
(A)∠A = 82.2°,∠C = 62.8°, c = 17.1
Step-by-step explanation:
In Triangle ABC
∠B=35°
a=19
b=11
Using Law of SInes
[tex]\dfrac{a}{\sin A} =\dfrac{b}{\sin B} \\\dfrac{19}{\sin A} =\dfrac{11}{\sin 35^\circ} \\11*\sin A=19*\sin 35^\circ\\\sin A=(19*\sin 35^\circ) \div 11\\A= \arcsin [(19*\sin 35^\circ) \div 11]\\A=82.2^\circ[/tex]
Now:
[tex]\angle A+\angle B+\angle C=180^\circ\\35^\circ+82.2^\circ+\angle C=180^\circ\\\angle C=180^\circ-[35^\circ+82.2^\circ]\\\angle C=62.8^\circ[/tex]
Using Law of Sines
[tex]\dfrac{c}{\sin C} =\dfrac{a}{\sin A} \\\dfrac{c}{\sin 62.8^\circ} =\dfrac{19}{\sin 82.2^\circ}\\c=\dfrac{19}{\sin 82.2^\circ}*\sin 62.8^\circ\\\\c=17.1[/tex]
Therefore:
∠A = 82.2°,∠C = 62.8°, c = 17.1
The correct option is A.
Answer:
it is a
∠A = 82.2°,∠C = 62.8°, c = 17.1
Step-by-step explanation:
give the other dude brainliest he is correct
