Question

Solve the triangle. A = 52°, b = 14, c = 6 a ≈ 14.9, C ≈ 24.2, B ≈ 103.8 a ≈ 11.3, C ≈ 24.2, B ≈ 103.8 a ≈ 14.9, C ≈ 28.2, B ≈ 99.8 No triangles possible

Answer 1

If you consider

Option (A) in which

b = 14, c = 6, a = 14.9→→→→Sides of Triangle

A= 52°, C = 24.2°, B = 103.8°→→Angles of Triangle

As , you can see that sum of two sides is greater than third side

And , sum of three angles of triangle is 180°.

b +c= 14 +6=20>a=14.9

c+a=6+14.9=20.9>b=14

a+b=14.9 +14=28.9>c=6

Sum of the angles = (52)° + (24.2)°+ (103.8)°

= (52)° + (128)°

= 180°

So, yes ,triangle is Possible.

Answer 2

B≈103.8° , a≈11.3 and C≈24.2°

B is correct.

Explanation:

Given:
`A=52^\circ, b=14 \text{ and } c=6`

Using sine find rest part of the triangle.

Sine law:

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

Cosine Law:

`a^2=b^2+c^2-2bc\cos A`

where,
`A=52^\circ, b=14 \text{ and } c=6`

`(a)/(\sin 52^\circ)=(14)/(\sin B)=(6)/(\sin C)`

`a^2=14^2+6^2-2(14)(6)\cos 52^\circ=158.57`

`a\approx 11.3`

`(a)/(\sin A)=(b)/(\sin B)`

`(11.3)/(\sin 52^\circ)=(6)/(\sin B)`

`\sin B=0.4184`

`B\approx 103.8`

`A+B+C=180^\circ`

`C\approx 180-52-103.8 = 24.2`

Hence, B≈103.8° , a≈11.3 and C≈24.2°