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

Option A is correct.

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.

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

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

`(6)/(sin\,A)=(14.1)/(sin\,109)`

`sin\,A=(0.95)/(14.1)*6`

`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.