Question

The number of triangles ABC that can be formed with a=3,b=8 and $\sin A=\dfrac{5}{13}$ is [Roorkee Qualifying 1998]
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint:
In this question, we are provided with the two sides and sine measurement of one angle. In order to find the number of triangles formed we will apply the law of the sines formula, relating the given lengths of the sides of the triangle to the sines of their consecutive angles.

Formula Used:
Laws of Sines for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$\dfrac{a}{\sin A}= \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} =\dfrac{\sin C}{c}$

Complete step-by-step solution:
It is given that in the triangle $ABC$, $a=3,b=8$ and $\sin A=\dfrac{5}{13}$.
Applying the law of sines formula we get;
$\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$
$\Rightarrow\dfrac{\sin B}{8}=\dfrac{5/13}{3}$
$\Rightarrow sin B=\dfrac{8}{3}\times\dfrac{5}{13}$
\[\Rightarrow \sin B=\dfrac{40}{39}>1\]
Since the value is greater than $1$ it is not possible. So, no triangle is possible with the $a=3,b=8$ and $\sin A=\dfrac{5}{13}$.

So, option A is correct.

Note:
The alternative way to find the numbers of triangles formed we can use the law of cosines. Both the law of sines and cosines are applicable in finding the unknown values of the angle or an unknown side of a given triangle. Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$a^2 = b^2 + c^2 − 2bc.\cos A$
$b^2 = a^2 +c^2 − 2ac.\cos B$
$c^2 = a^2 + b^2 − 2ab.\cos C$.