Chapter 9.7, Problem 24PSC

To determine

To calculate: The Values of x and y.

Answer to Problem 24PSC

The Values of x and y are $3\sqrt{3}$ and $\frac{3\sqrt{3}}{2}$ respectively.

Explanation of Solution

Given information:Here, we have given a triangle.

Formula used: Sum of all interior angles of a triangle is $180°$ .

Calculation: Let the triangle be $\Delta ABC$ with a point D on AB . Draw a perpendicular from D to the base BC such that $DP\perp BC$ .

In $\Delta ABC$ ,

  $\begin{array}{l}\angle A+\angle B+\angle C=180°\\ \Rightarrow 90°+30°+\angle C=180°\\ \Rightarrow 120°+\angle C=180°\\ \Rightarrow \angle C=180°-120°\\ \Rightarrow \angle C=60°\end{array}$

Now, $\angle DCB=\frac{\angle C}{2}$

  $\begin{array}{l}\Rightarrow \angle DCB=\frac{60°}{2}\\ \Rightarrow \angle DCB=30°\end{array}$

Now, in $\Delta DBC$ ,

  $\begin{array}{l}\angle BDC+\angle BCD+\angle C=180°\\ \Rightarrow \angle BDC+30°+30°=180°\\ \Rightarrow \angle BDC+60°=180°\\ \Rightarrow \angle BDC=180°-60°\\ \Rightarrow \angle BDC=120°\end{array}$

Now, $\angle PDB=\frac{\angle BDC}{2}$ (Perpendicular bisects the angle)

  $\begin{array}{l}\Rightarrow \angle PDB=\frac{120°}{2}\\ \Rightarrow \angle PDB=60°\end{array}$

Now, In $\Delta PDB$ ,

Angles are $90°,60°,30°$

  $\Rightarrow$ Hypotenuse = $2a$ and perpendicular and base will be $a$ and $a\sqrt{3}$ respectively.

  $\begin{array}{l}\Rightarrow 3=2a\\ \Rightarrow a=\frac{3}{2}\end{array}$

So, Base ( $a\sqrt{3}$ ) = $\frac{3}{2}(\sqrt{3})$

  $=\frac{3\sqrt{3}}{2}$

Now, BC = BP + CP and BP = CP (Perpendicular bisects the base)

BC = 2(BP)

BC = $(2)\frac{3\sqrt{3}}{2}$

BC = $3\sqrt{3}$

Now, in $\Delta ABC$ ,

BC (y) = Base $=\frac{3\sqrt{3}}{2}$