Chapter 9.6, Problem 21E

To determine

Tofind:The value of $x$

Answer to Problem 21E

  $x=15\sqrt{2}$

Explanation of Solution

Given information:

  ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ right angled triangle

Length of hypotenuse of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle $=10\sqrt{3}$

Concept Used:

In a ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ right angled triangle, length of both legs are equaland length of hypotenuse is given by product of length of any leg and square root of 2

In a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ right angled triangle, length of longer leg is product of shorter length and square root of 3 and length of hypotenuse is given by twice the length of shorter leg.

Calculation:

Let length of shorter leg of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ be $a$

Length of hypotenuse is twice the length of shorter leg-

Therefore

  $\begin{array}{l}10\sqrt{3}=2a\\ \Rightarrow a=5\sqrt{3}\end{array}$

Letlength of longer legof ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ be $b$

Length of longer leg is product of shorter length and square root of 3

Therefore

  $\begin{array}{l}b=a\sqrt{3}\\ \Rightarrow b=5\sqrt{3}\left(\sqrt{3}\right)\\ \Rightarrow b=5\left(3\right)\\ \Rightarrow b=15\end{array}$

Longer length of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle is also leg length of ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ triangle with hypotenuse $x$

Length of hypotenuse in ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ product of length of any leg and square root of 2

Therefore-

  $x=15\sqrt{2}$