Question

1. calculate the missing angles and the slope of the hypotenuse. round to the nearest tenth.

a.
b.
c.
d.

Explanation:

Step1: Recall triangle - angle sum property

The sum of angles in a right - triangle is 180°. One angle is 90°. Let the other non - right angles be \(A\) and \(B\), so \(A + B+90^{\circ}=180^{\circ}\), or \(A + B = 90^{\circ}\).

Step2: Recall Pythagorean theorem for hypotenuse

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(c=\sqrt{a^{2}+b^{2}}\). The slope of the hypotenuse in a right - triangle with legs \(a\) and \(b\) (where \(a\) is the horizontal leg and \(b\) is the vertical leg) is \(m=\frac{b}{a}\) (for the non - vertical hypotenuse).

Case c:

Let the legs of the right - triangle be \(a = 19\) m and \(b = 26\) m.

Find the missing non - right angles:

Let one non - right angle be \(\theta\). \(\tan\theta=\frac{26}{19}\), so \(\theta=\arctan(\frac{26}{19})\approx53.8^{\circ}\). The other non - right angle is \(90^{\circ}-\theta\approx90^{\circ}-53.8^{\circ}=36.2^{\circ}\).

Find the hypotenuse:

\(c=\sqrt{19^{2}+26^{2}}=\sqrt{361 + 676}=\sqrt{1037}\approx32.2\) m. The slope of the hypotenuse is \(\frac{26}{19}\approx1.4\).

Case d:

Let the legs of the right - triangle be \(a = 16\) ft and \(b = 22\) ft.

Find the missing non - right angles:

Let one non - right angle be \(\alpha\). \(\tan\alpha=\frac{22}{16}=\frac{11}{8}\), so \(\alpha=\arctan(\frac{11}{8})\approx53.9^{\circ}\). The other non - right angle is \(90^{\circ}-\alpha\approx90^{\circ}-53.9^{\circ}=36.1^{\circ}\).

Find the hypotenuse:

\(c=\sqrt{16^{2}+22^{2}}=\sqrt{256+484}=\sqrt{740}\approx27.2\) ft. The slope of the hypotenuse is \(\frac{22}{16}=1.4\).

Case b:

Let the legs of the right - triangle be \(a = 102\) mm and \(b = 58\) mm.

Find the missing non - right angles:

Let one non - right angle be \(\beta\). \(\tan\beta=\frac{58}{102}\), so \(\beta=\arctan(\frac{58}{102})\approx29.9^{\circ}\). The other non - right angle is \(90^{\circ}-\beta\approx90^{\circ}-29.9^{\circ}=60.1^{\circ}\).

Find the hypotenuse:

\(c=\sqrt{102^{2}+58^{2}}=\sqrt{10404 + 3364}=\sqrt{13768}\approx117.3\) mm. The slope of the hypotenuse is \(\frac{58}{102}\approx0.6\).

Answer:

Case c:

Missing angles: approximately \(53.8^{\circ}\) and \(36.2^{\circ}\), Hypotenuse: approximately \(32.2\) m, Slope of hypotenuse: approximately \(1.4\)

Case d:

Missing angles: approximately \(53.9^{\circ}\) and \(36.1^{\circ}\), Hypotenuse: approximately \(27.2\) ft, Slope of hypotenuse: approximately \(1.4\)

Case b:

Missing angles: approximately \(29.9^{\circ}\) and \(60.1^{\circ}\), Hypotenuse: approximately \(117.3\) mm, Slope of hypotenuse: approximately \(0.6\)