Question

Solve the triangle. A = 51°, b = 11, c = 7

Answer 1

There are no triangles possible here!

Explanation:

Answer 2

Answer with explanation:

The Description of triangle is :

A=51°, b=11 and c=7

We will use cosine law to find length of third side

`\cos A=(b^2+c^2-a^2)/(2* b * c)\\\\ \ cos 51^(\circ)=(11^2+7^2-a^2)/(2* 11 * 7)\\\\0.63=(121+49-a^2)/(154)\\\\ 154 * 0.63=170-a^2\\\\97.02=170 -a^2\\\\a^2=170 -97.02\\\\a^2=72.98\\\\a=8.55`

a=8.55(approx)

Now, we will use Sine law to find other angle.

`(a)/(\sin A^(\circ))=(b)/(\sin B^(\circ))\\\\ (8.55)/(\sin 51^(\circ))=(11)/(\ sinB^(\circ))\\\\ \ sinB^(\circ)=(11 *0.78)/(8.55)\\\\\sin B^(\circ)=(8.55)/(8.55) \\\\ sinB^(\circ)=1\\\\B=90^(\circ)`

We will use angle sum property of triangle to find measure of angle C .

→∠A + ∠B+∠C=180°

→ 51°+90°+∠C=180°

→∠C=180° - 141°

→∠C=39°

Side Opposite to Angle A= 8.55 unit

∠ B=90°

∠ C=39°