Question

In the given diagram, ABCD is a square and $\Delta BCT$ is an equilateral triangle. $\angle BTD$equals

a) ${{30}^{\circ }}$
b) ${{15}^{\circ }}$
c) ${{45}^{\circ }}$
d) ${{35}^{\circ }}$

Answer 1

Hint: Since it is given that ABCD is a square and $\Delta BCT$ is an equilateral triangle. Use the properties of the square and equilateral triangle to find the relation between angles sides of the square and triangle. Then find $\angle BTD$.

Complete step-by-step answer:
As we know that ABCD is a square. Therefore, all the angles of the square are equal to ${{90}^{\circ }}$ .
So, we can say that $\angle BCD={{90}^{\circ }}$
Also, $\Delta BCT$ is an equilateral triangle. Therefore, all the angles of the equilateral triangle are equal to ${{60}^{\circ }}$.
So, we can say that $\angle BCT={{60}^{\circ }}$
Now, consider $\Delta TDC$ in the following figure.

As we know the sum of all angles of a triangle is ${{180}^{\circ }}$. So, we can say that:
$\angle TDC+\angle TCD+\angle DTC={{180}^{\circ }}......(1)$
As we know that BC is the common side of the square and the equilateral triangle. Therefore, we can say that:
CD = CT
In triangle $\Delta TDC$;
Since CD = CT, so $\Delta TDC$ is an isosceles triangle.
This implies that $\angle DTC=\angle TDC$
Let us assume that, $\angle TDC=\angle TCD=x$
So, we can write equation (1) as:
$\begin{align}
  & \Rightarrow x+x+\angle TCD={{180}^{\circ }} \\
 & \Rightarrow 2x+\angle TCD={{180}^{\circ }}......(1) \\
\end{align}$
Also,
$\begin{align}
  & \angle TCD=\angle BCD+\angle BCT \\
 & ={{90}^{\circ }}+{{60}^{\circ }} \\
 & ={{150}^{\circ }}.....(2)
\end{align}$
Substitute equation (2) in equation (1), we get:
$\begin{align}
  & \Rightarrow 2x+{{150}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow 2x={{30}^{\circ }} \\
 & \Rightarrow x={{15}^{\circ }} \\
\end{align}$
Hence, $\angle DTC={{15}^{\circ }}$
Now, in $\Delta BCT$, all the angles are equal to ${{60}^{\circ }}$
So, $\angle BTC={{60}^{\circ }}......(3)$
We can write equation (3) as:
$\begin{align}
  & \Rightarrow \angle BTD+\angle DTC={{60}^{\circ }} \\
 & \Rightarrow \angle BTD+{{15}^{\circ }}={{60}^{\circ }} \\
 & \Rightarrow \angle BTD={{45}^{\circ }} \\
\end{align}$

So, the correct answer is “Option c”.

Note: There is another way to solve for $\angle BTD$
Since, we have found that $\angle TDC={{15}^{\circ }}$.
Let BC and TD intersect at point E as shown in the figure below. So, we have $BC\parallel AD$.

Therefore, $\angle ADE=\angle BET={{\left( 90-15 \right)}^{\circ }}$ [corresponding angles]
So, we have: $\angle BET={{75}^{\circ }}$
Now, in $\Delta BET$
$\begin{align}
  & \Rightarrow \angle BTE+\angle BET+\angle TBE={{180}^{\circ }} \\
 & \Rightarrow \angle BTE+{{75}^{\circ }}+{{60}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow \angle BTE={{45}^{\circ }} \\
\end{align}$