Question

Solve the triangle.

A = 52°, b = 14, c = 6

Answer 1

Answer:

B=103.8° , a=11.3 and C=24.2°

Given:

A = 52°, b = 14, c = 6

using sine law:

${a\over sinA}={b\over sinB}={c\over sinC}$

${a\over sin52^o}={14\over sinB}={6\over sinC}$ ..........................1

using cosine law:

$a^2=b^2+c^2-2bcCosA$

$a^2=196+36-2(14)(6)Cos52^o$ .............2

from 2 we get,

a=11.3 ........................3

putting 3 in 1

And solving

${11.3\over sin52^o}={14\over sinB}$

And

${11.3\over sin52^o}={6\over sinC}$

As we see first possible solution

we get

B=103.8⁰

C=24.2⁰