Question

How do you solve the right angle ABC (where C=90 degree) when A=31.5 degree, b=29.7ft ?

Answer 1

Hint: Subtract the sum of angles A and C from 180 degree to get the value of angle B. Put the values of angle B and side ‘b’ in $\tan B=\dfrac{b}{a}$ to get the value of ‘a’. Again put the values of angle B and side ‘a’ in $\cos B=\dfrac{a}{c}$ to get the value of ‘c’.

Complete step by step solution:

ABC is a right angle triangle with base ‘a’, perpendicular ‘b’ and hypotenuse ‘c’.
Given, in right angle triangle ABC angle A=31.5 degree and angle C=90 degree
As we know the sum of the angles of a triangle is 180 degree
So, angle ‘B’ will be $=180-\left( 31.5+90 \right)=180-121.5=58.5$ degree
In right angle triangle ABC, $\tan B=\dfrac{AC}{BC}=\dfrac{b}{a}$
Putting the values of angle B and side ‘b’, we get
$\begin{align}
  & \Rightarrow \tan \left( 58.5 \right)=\dfrac{29.7}{a} \\
 & \Rightarrow 1.63=\dfrac{29.7}{a} \\
 & \Rightarrow a=\dfrac{29.7}{1.63} \\
 & \Rightarrow a=18.22 \\
\end{align}$
Again in right angle triangle ABC, $\cos B=\dfrac{BC}{AB}=\dfrac{a}{c}$
Putting the values of angle B and side a, we get
$\begin{align}
  & \Rightarrow \cos \left( 58.5 \right)=\dfrac{18.22}{c} \\
 & \Rightarrow 0.52=\dfrac{18.22}{c} \\
 & \Rightarrow c=\dfrac{18.22}{0.52} \\
 & \Rightarrow c=34.87 \\
\end{align}$
From the above, we conclude that in triangle ABC the angles are $\angle A={{31.5}^{\circ }}$ , $\angle B={{58.5}^{\circ }}$, $\angle C={{90}^{\circ }}$ and the sides are a=18.22 ft, b=29.7 ft, c=34.87 ft.
This is the required solution.

Note: For the above triangle, Pythagoras' theorem can be applied as ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$.
Now, we have a=18.22 and b=29.7
Using Pythagoras' theorem and putting the values of ‘a’ and ‘b’, we get
$\begin{align}
  & {{c}^{2}}={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{c}^{2}}={{\left( 18.22 \right)}^{2}}+{{\left( 29.7 \right)}^{2}} \\
 & \Rightarrow {{c}^{2}}=331.96+882.09 \\
 & \Rightarrow c=\sqrt{1214.06} \\
 & \Rightarrow c=34.87 \\
\end{align}$
This is the alternative method.