Question

If A=${30^ \circ }$, a=7, b=8 in △ABC, then prove that B has two solutions.

Answer 1

Hint: In order to solve this problem we need to use sine rule and solve. We know the formula that \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}\], where a, b are sides of triangles and A, B are angles of triangle opposite to side a and b.



Complete step-by-step answer:

Above figure is the diagram of the triangle where, AB = c, BC=a=7 and AC = b=8, angle A = 30 degrees (given).
As from sine rule,
We know that in the triangle ABC,
\[ \Rightarrow \dfrac{{\text{a}}}{{{\text{sinA}}}}{\text{ = }}\dfrac{{\text{b}}}{{{\text{sinB}}}}\]
Above equation can also be written as:

$ \Rightarrow {\text{sinB = }}\dfrac{{\text{b}}}{{\text{a}}}{\text{sinA}}$
The value of a, b, angle A is given as above. So,
$ \Rightarrow {\text{sinB = }}\dfrac{{\text{b}}}{{\text{a}}}{\text{sinA}} = \dfrac{{\text{8}}}{{\text{7}}}{\text{ \times }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = }}\dfrac{{\text{4}}}{{\text{7}}}$
​As the value of sin is same between (0,π), therefore, we have two solution $ \Rightarrow B = {\sin ^{ - 1}}\left( {\dfrac{4}{7}} \right)\,\,\,\,\,$ (1st quadrant)
${\text{and }}\,\,\,\pi - {\sin ^{ - 1}}\left( {\dfrac{4}{7}} \right)$ (2nd quadrant $\because \,\sin (\pi - \theta ) = \sin \theta $)
Hence, the two solutions of B can be written as:
$B = {\sin ^{ - 1}}\left( {\dfrac{4}{7}} \right)\,\,\,\,{\text{and}}\,\,\,\pi - {\sin ^{ - 1}}\left( {\dfrac{4}{7}} \right)$

Note:- In this problem we have to use the sine rule that is \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}\] and need to know about the quadrant in which quadrant what is the value of angle. Proceeding with this you will get the correct answer.