In a right angle triangle ABC, the angle C = $90^{\circ}$, A = $45^{\circ}$ and BC = 7 units. Find the remaining angles and sides.

Hint:First we will draw the required diagram and then we will use the fact that the sum of all the angles in a triangle is 180, with that we will find the value of $\angle B$ . And then we will use the formula of sin or tan to find the remaining sides.


Complete step-by-step answer:

In the above diagram a = BC = 7.
Now it is given $\angle A$ = 45, now we will use the fact that the sum of all the angles in a triangle is 180.
Hence we get,
$\angle A+\angle B+\angle C=180$
Substituting the values of $\angle A$ = 45 and $\angle C=90$ we get,
$\begin{align}
  & \angle B+45+90=180 \\
 & \angle B=180-90-45 \\
 & \angle B=45 \\
\end{align}$
Now we know that,
$\begin{align}
  & \sin B=\dfrac{height}{hypotenuse} \\
 & \tan B=\dfrac{height}{base} \\
\end{align}$
Now substituting the values of B = 45, height = b, hypotenuse = c, and base = a = 7 we get,
We know that tan45 = 1,
$\begin{align}
  & \tan 45=\dfrac{b}{a} \\
 & b=7 \\
\end{align}$
Now in the formula of sin using b = 7 we get,
We know that $\sin 45=\dfrac{1}{\sqrt{2}}$
$\begin{align}
  & \sin B=\dfrac{b}{c} \\
 & c=\dfrac{7}{\sin 45} \\
 & c=7\sqrt{2} \\
\end{align}$
Hence, we have all the values that has been asked in the question.

Note: To solve this question one can also use the cosine formula $\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}$ and sin formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$ , and then we will have to substitute the values in these two formula to find the value of the other two sides of the triangle. One can also use Pythagoras theorem ${{b}^{2}}={{c}^{2}}-{{a}^{2}}$ to find the value of third side given that the other two sides are known.