Question

In the adjoining figure,$\Delta ABC$ is right-angled at B and $\angle A={{45}^{\circ }}$. If \[AC=3\sqrt{2}\text{ cm}\], find
(i) BC
(ii) AB.

Answer 1

Hint: Assume AB as base and BC as perpendicular of the given right-angle triangle. So, AC must be the hypotenuse. Use the formula of the trigonometric functions given by: $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$ and substitute the value of $\theta $ equal to 45 degrees, to determine the value of base AB. Once the value of the base AB is determined, use Pythagoras theorem given by: $\text{hypotenus}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicula}{{\text{r}}^{\text{2}}}$ to determine the value of perpendicular BC.

Complete step by step answer:

From the figure, we can see that $\angle A={{45}^{\circ }}$ is given to us. So, the side lying opposite to it is considered as perpendicular. Therefore, BC is perpendicular. Hence, AB must be the base.
Here, AC is considered as the hypotenuse because it is lying opposite to the 90 degree angle.
Now, let us come to the question. we have to determine the value of AB and BC.
In right angle triangle ABC,
Using the formula: $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$, where, $\theta $ is the angle A = 45 degrees, we get,
$\cos {{45}^{\circ }}=\dfrac{AB}{AC}$
Substituting, $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ and \[AC=3\sqrt{2}\text{ cm}\], we get,
$\begin{align}
  & \dfrac{1}{\sqrt{2}}=\dfrac{AB}{3\sqrt{2}} \\
 & \Rightarrow AB=\dfrac{3\sqrt{2}}{\sqrt{2}} \\
 & \Rightarrow AB=3\text{ cm} \\
\end{align}$
 Now, using Pythagoras theorem: $\text{hypotenus}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicula}{{\text{r}}^{\text{2}}}$, we have,
$\begin{align}
  & \text{perpendicula}{{\text{r}}^{\text{2}}}=\text{hypotenus}{{\text{e}}^{\text{2}}}-\text{bas}{{\text{e}}^{\text{2}}} \\
 & \text{perpendicular}=\sqrt{\text{hypotenus}{{\text{e}}^{\text{2}}}-\text{bas}{{\text{e}}^{\text{2}}}} \\
 & \Rightarrow BC=\sqrt{{{\left( 3\sqrt{2} \right)}^{\text{2}}}-{{3}^{\text{2}}}} \\
 & \Rightarrow BC=\sqrt{18-9} \\
 & \Rightarrow BC=\sqrt{9} \\
 & \Rightarrow BC=3\text{ cm} \\
\end{align}$
(i) BC = 3 cm
(ii) AB = 3 cm
Note: We can clearly see that the triangle is an isosceles right angle triangle. It was provided to us in the question that, $\Delta ABC$ is right-angled at B and $\angle A={{45}^{\circ }}$. From this information only we can say that the triangle is also an isosceles triangle because when we will find the third angle of the triangle, that is angle C, then it will be 45 degrees. Therefore, the alternate method of solving this question is that we can assume the length of equal sides as ‘x’ and apply the Pythagoras theorem in the first step only to determine the value of ‘x’. In the above solution, we have used cosine of the given angle, you may also use sine of the given angle. The answer will be the same.