Question

If \[a = 2\sqrt 2 ,b = 6,A = {45^0}\], then
A. no triangle is possible
B. one triangle is possible
C. two triangles are possible
D. either no triangle is possible, or two triangles are possible

Answer 1

Hint: This is a question based on the properties of a triangle. Basic concepts like the sine rule and the range of the sine function are used to solve this question. The Law of Sines (or Sine Rule) is very useful for solving triangles: \[\dfrac{a}{{\sin (A)}} = \dfrac{b}{{\sin (B)}} = \dfrac{c}{{\sin (C)}}\], here $a, b, c$ are the sides of the triangles and $A, B, C$ are the angles opposite to these sides.

Complete step by step answer:
Now, the range of sine function is given by:
\[ - 1 \leqslant \sin (\theta ) \leqslant 1\]
where\[\theta \] is any real number.
In this question, we are given two sides and an angle, so if they satisfy the properties of a triangle, then the sine rule is applicable,
\[ \Rightarrow \dfrac{a}{{\sin (A)}} = \dfrac{b}{{\sin (B)}} = \dfrac{c}{{\sin (C)}}\]
Now, we are given two sides and an angle, so let us find the other angle
\[ \Rightarrow \dfrac{a}{{\sin (A)}} = \dfrac{b}{{\sin (B)}}\]
Let us put the given values of the sides and angle in the formula
\[ \Rightarrow \dfrac{{2\sqrt 2 }}{{\sin ({{45}^0})}} = \dfrac{6}{{\sin (B)}}\]

Now, putting the value of \[\sin ({45^0}) = \dfrac{1}{{\sqrt 2 }}\]we get,
\[ \Rightarrow \dfrac{{2\sqrt 2 }}{{\dfrac{1}{{\sqrt 2 }}}} = \dfrac{6}{{\sin (B)}}\]
Clearly, the denominator will be multiplied by the numerator
\[ \Rightarrow 2\sqrt 2 \times \sqrt 2 = \dfrac{6}{{\sin (B)}}\]
Now, simplifying the left-hand side we get,
\[ \Rightarrow 4 = \dfrac{6}{{\sin (B)}}\]
Cross multiplying the equation we get,
\[ \Rightarrow \sin (B) = \dfrac{6}{4}\]
Now, simplifying the right-hand side we get,
\[ \Rightarrow \sin (B) = \dfrac{3}{2}\]
The value of sine function can never be greater than one; therefore, we get,
\[ \Rightarrow \sin (B) > 1\]
Thus, no triangle is possible.

Hence, option A is correct.

Note: This question involves the concepts of properties of triangles and trigonometry. One should be well versed with these concepts to solve this question. Do not commit calculation mistakes, and be sure of the final answer.In Trigonometry, different types of problems can be solved using trigonometry formulas. These problems may include trigonometric ratios (sin, cos, tan, sec, cosec and cot), Pythagorean identities, product identities, etc.