Question

Solve the triangle given that a=19 b=16, c=11.

Answer 1

Answer:

The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.

Step-by-step explanation:

From statement we know all sides of the triangle ($a$, $b$, $c$), but all angles are unknown ($A$, $B$, $C$). (Please notice that angles with upper case letters represent the angle opposite to the side with the same letter but in lower case) From Geometry it is given that sum of internal angles of triangles equal 180º, we can obtain the missing information by using Law of Cosine twice and this property mentioned above.

If we know that $a = 19$, $b = 16$ and $c = 11$, then the missing angles are, respectively:

Angle A

$a^{2} = b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A$ (1)

$A = \cos^{-1}\left(\frac{b^{2}+c^{2}-a^{2}}{2\cdot b\cdot c} \right)$

$A = \cos^{-1}\left[\frac{16^{2}+11^{2}-19^{2}}{2\cdot (16)\cdot (11)} \right]$

$A \approx 87.395^{\circ}$

Angle B

$b^{2} = a^{2}+c^{2}-2\cdot a\cdot c \cdot \cos B$ (2)

$B = \cos^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2\cdot a\cdot c} \right)$

$B = \cos^{-1}\left[\frac{19^{2}+11^{2}-16^{2}}{2\cdot (19)\cdot (11)} \right]$

$B\approx 57.271^{\circ}$

Angle C

$C = 180^{\circ}-A-B$

$C = 180^{\circ}-87.395^{\circ}-57.271^{\circ}$

$C = 35.334^{\circ}$

The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.