Perform the calculation. \( 90^{\circ}-53^{\circ} 40^{\prime} 31^{\prime \prime} \) \( 90^{\circ}-53^{\circ} 40^{\prime} 31^{\prime \prime}=\square^{\circ} \square^{\prime \prime} \) (Simplify your answers. Type an integer or a fraction.)

To perform the calculation, we need to subtract \( 53^{\circ} 40^{\prime} 31^{\prime\prime} \) from \( 90^{\circ} \). 1. First, convert \( 90^{\circ} \) into degrees, minutes, and seconds format: \( 90^{\circ} = 90^{\circ} 0^{\prime} 0^{\prime\prime} \). 2. Now, we align it for subtraction: \[ \begin{array}{c@{}c@{}c@{}c@{}c@{}c@{}c@{}c} & 90^{\circ} & 0^{\prime} & 0^{\prime\prime} \\ - & 53^{\circ} & 40^{\prime} & 31^{\prime\prime} \\ \end{array} \] 3. Start with the seconds: \[ 0^{\prime\prime} - 31^{\prime\prime} \quad \text{(we can't do this, so we borrow)} \] We borrow 1 minute (which is \( 60^{\prime\prime} \)), so: \[ 60^{\prime\prime} - 31^{\prime\prime} = 29^{\prime\prime} \] Now we have \( 90^{\circ} 0^{\prime} - 1^{\prime} \) remaining from our borrowing: \[ 90^{\circ} 0^{\prime} \rightarrow 89^{\circ} 59^{\prime} \] 4. Now subtract the minutes: \[ 59^{\prime} - 40^{\prime} = 19^{\prime} \] 5. Finally, subtract the degrees: \[ 89^{\circ} - 53^{\circ} = 36^{\circ} \] Combining everything, we get: \[ 36^{\circ} 19^{\prime} 29^{\prime\prime} \] Thus, \[ 90^{\circ}-53^{\circ} 40^{\prime} 31^{\prime\prime} = 36^{\circ} 19^{\prime} 29^{\prime\prime} \]