Question

Given A= 25 B= 43 and C=17, solve triangle ABC. Round to the nearest hundredth.

Answer 1

I assume  A, B and C are the side lengths
You can find the values of the angles using the cosine rule.

Answer 2

Answer:

It is given that , In ΔABC , in which, ∠A=25°, ∠B=43°,

Length of side c= 17 units

As, angle sum property of triangle says that

∠A +∠B +∠C=180°

25°+43°+ ∠C=180°

∠C=180° -68°

∠C=112°

Using Sine Rule

$\frac{a}{Sin A^{\circ}}=\frac{b}{Sin B^{\circ}}=\frac{c}{Sin C^{\circ}}\\\\ \frac{a}{Sin 25^{\circ}}=\frac{b}{Sin 43^{\circ}}=\frac{17}{Sin 112^{\circ}}\\\\ a=\frac{17 \times Sin 25^{\circ}}{Sin 112^{\circ}}\\\\a=\frac{17 \times 0.4226}{0.9271}\\\\a=\frac{7.1842}{0.9271}\\\\a=7.749=7.75\\\\ b=\frac{17 \times Sin 43^{\circ}}{Sin 112^{\circ}}\\\\b=\frac{17 \times 0.6819}{0.9271}\\\\b=\frac{11.5923}{0.9271}\\\\b=12.5038=12.50$

So, Side opposite to angle A=a=7.75 units

Side opposite to angle B=b=12.50 units

And , ∠C=112°