Question

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 6, c = 11, B = 109° A. b = 14.1, A = 24°, C = 47° B. b = 19.9, A = 22°, C = 49° C. b = 17, A = 26°, C = 45° D. no triangle

Answer 1

Use cosine rule to find  b

b^2 = 6^2 + 11^2 - 2*6*11cos 109  =   199.97

b =  14.1  to nearest tenth.

now use sine rule to find the value of < A

14.14 / sin 109 = 6 /  sin A

<A =  24 degrees

<C = 180 - 24 - 109  = 47 degrees

Its  A

Answer 2

Answer:

Option A is correct.

Step-by-step explanation:

Given: a = 6 , c = 11 ,  B = 109°

To find : b , ∠A , ∠C

We use Law of cosine and Law of sines.

Law of Cosine used for calculating one side of a triangle when the angle opposite and the other two sides are known.

b² = a² + c² - 2ac.cos B

Law of Sines is a rule relating the sides and angles of any triangle.

$\frac{a}{sin\,A}=\frac{b}{sin\,B}=\frac{c}{sin\,C}$

Using Law of Cosines, we get

b² = 6² + 11² - 2 × 6 × 11 × cos 109

b² = 199.97

b = 14.1 (nearest tenth)

Now using Law of sines, we get

$\frac{a}{sin\,A}=\frac{b}{sin\,B}$

$\frac{6}{sin\,A}=\frac{14.1}{sin\,109}$

$sin\,A=\frac{0.95}{14.1}\times6$

$A=sin^{-1}\,(0.404)$

∠ A = 23.8° = 24° (nearest degree)

Using Angle sum property of triangle,

∠A + ∠B + ∠C = 180°

∠C = 180 - (24 + 109)

∠C = 47°

Therefore, Option A is correct.