Question

Find the measure of angle B for the triangle with the following dimensions: \[A = {45^ \circ },\] $ a = 60.5, $ $ b = 90 $ ?

Answer 1

Hint: In such questions, we are required to find the measure of angle B in the triangle with the dimensions given to us. There can be various methods by which we can find the value of an angle in a triangle. For such a question, we can make use of the sine rule to find the measure of the required angle.

Complete step by step solution:
In the given question, the required angle can be found using the sine rule of trigonometry. According to the sine rule of trigonometry, the ratio of sine of two angles is equal to the ratio of the sides opposite to the angles.
Using sine formula of $ \vartriangle ABC $ with standard expression $ \dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} $ and substituting the values of the angles and sides given to us in the question, we get,
 $ \Rightarrow \dfrac{{\sin {{45}^ \circ }}}{{60.5}} = \dfrac{{\sin B}}{{90}} $
We know that $ \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} $ . Hence, we get,
 $ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{60.5}} = \dfrac{{\sin B}}{{90}} $
 $ \Rightarrow \dfrac{{90}}{{60.5 \times \sqrt 2 }} = \sin B $
 $ \Rightarrow \dfrac{{{\text{1}}{\text{.4876033057851239669421487603306}}}}{{\sqrt 2 }} = \sin B $
So, the value of sine of angle B can be obtained by simplifying the expression given above.
On solving, we get $ \sin B > 1 $ which could not be possible as $ - 1 < \sin x < 1 $ for all real values of x.
Hence measure of B cannot be determined.

Note: We know that the value of sine and cosine functions always range between $ - 1 $ and $ 1 $ . Hence, there exists no value of angle x for which the value of sine function is greater than $ 1 $ or less than $ - 1 $ .