Question

In △ABC, m∠A=15°, a=9, and b=12. Find c to the nearest tenth.

Answer 1

Answer:

20.0 unit ( approx )

Step-by-step explanation:

Here,

ABC is a triangle in which,

m∠A=15°, a=9, and b=12,

By the law of sine,

$\frac{sin A}{a}=\frac{sin B}{b}=\frac{sin C}{c}----(1)$

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

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

By substituting the values,

$\implies sin B=\frac{12\times sin 15^{\circ}}{9}\approx 0.3451$

$\implies B \approx 20.19^{\circ}$

Now, by the property of triangle,

m∠A + m∠B+ m∠C = 180°

⇒ m∠C = 180° - 15° - 20.19° = 144.81°,

By the equation (1),

$c=\frac{b sin C}{sin B}=\frac{12\times sin 144.81^{\circ}}{sin 20.19^{\circ}}=20.0370532419\approx 20.0$

Answer 2

Answer:

=20.0

Step-by-step explanation:

We can first find the angle B using the sine rule as follows:

a/Sin A=b/Sin B

9/Sin 15=12/ Sin B

Sin B= (12 Sin 15)/9

=0.345

B=Sin⁻¹ 0.345

=20.18°

We then find C by using the summation of the interior angles of a triangle.

C=180-(20.18+15)

=144.82

Finding the length of c:

a/Sin A= c/ Sin C

9/Sin 15=c/Sin 144.82

c=(9 Sin 144.82)Sin 15

=20.0