Chapter 5.5, Problem 28PSC

To determine

To calculate: $\angle A$ in the given isosceles trapezoidTRAP.

Answer to Problem 28PSC

  $\angle A$ in the given isosceles trapezoid TRAPis ${150}^{\circ }59\text{'}06"$ .

Explanation of Solution

Given information: Given that TRAPis an isosceles trapezoid. Also,

  $\angle R=2.43+5.12\times \angle T$

…………………………. (i)

Concept used:

(a) Isosceles trapezoid has two equal smaller angles and two equal larger angles.

(b) Sum of all the angles of any quadrilateral is ${360}^{\circ }$ .

Calculation:

As TRAPis an isosceles trapezoid which means it has two equal smaller angles and two equal larger angles,

i.e., $\angle T=\angle P$ ………………………………. (ii)

  $\angle R=\angle A$ .……………………………… (iii)

We know that sum of all the angles of any quadrilateral is ${360}^{\circ }$ . So,

  $\angle T+\angle R+\angle A+\angle P={360}^{\circ }$ …………………. (iv)

Using equations (ii) and (iii) in equation (iv) to get,

  $\begin{array}{l}\angle T+\angle R+\angle R+\angle T={360}^{\circ }\\ 2\times \angle T+2\times \angle R={360}^{\circ }\end{array}$

Dividing the above equation by 2,

  $\angle T+\angle R={180}^{\circ }$ ……………………………. (v)

Using equation (i) in (v) to get,

  $\begin{array}{l}\angle T+2.43+5.12\times \angle T={180}^{\circ }\\ 6.12\times \angle T={177.57}^{\circ }\\ \angle T={29.015}^{\circ }\end{array}$

Now using the calculated value of $\angle T$ in equation (v),

  $\begin{array}{l}29.015+\angle R={180}^{\circ }\\ \angle R={150.985}^{\circ }\\ \end{array}$

As we know $1^{\circ }=60\text{'}$

So, ${0.985}^{\circ }=0.985\times 60\text{'}=59.1\text{'}$ …………………. (vi)

Also, $1\text{'}=60"$

then, $0.1\text{'}=0.1\times 60"=6"$ …………………... (vii)

from equations (vi) and (vii) we have,

  ${0.985}^{\circ }=59\text{'}06"$

Finally, we have $\angle R={150.985}^{\circ }={150}^{\circ }59\text{'}06"$

By using equation (iii) we get,

  $\angle A=\angle R={150}^{\circ }59\text{'}06"$

Hence, $\angle A$ in the given isosceles trapezoid TRAP is ${150}^{\circ }59\text{'}06"$ .