Question

solve the right triangle. round your answers to the nearest tenth. a = \\(\square^\circ\\) \\(a = \square\\) \\(b = \square\\) (triangle with hypotenuse 23, angle 46°, right angle, sides a, b, angle a)

Explanation:

Step1: Find angle A

In a right triangle, the sum of angles is \(180^\circ\). One angle is \(90^\circ\) (right angle), another is \(46^\circ\). So \(A = 180^\circ - 90^\circ - 46^\circ = 44^\circ\).

Step2: Find side a (adjacent to \(46^\circ\), hypotenuse is 23? Wait, no, wait. Wait, the side labeled 23 is the hypotenuse? Wait, no, looking at the triangle: the right angle is at the bottom, the angle of \(46^\circ\) is at the left, side opposite to \(A\) is \(a\)? Wait, no, let's clarify: in the triangle, the hypotenuse is 23? Wait, no, the side labeled 23 is the side opposite the right angle? Wait, no, the right angle is between \(a\) and \(b\), so the hypotenuse is the side labeled 23 (the one at the top, between the \(46^\circ\) angle and angle \(A\)). So hypotenuse \(c = 23\). Then, angle at left is \(46^\circ\), so:

To find \(a\) (opposite to angle \(A\)? Wait, no, \(a\) is adjacent to \(46^\circ\)? Wait, no, let's define: let's call the right angle \(C\), angle at left \(B = 46^\circ\), angle at \(A\) is \(A\), hypotenuse \(c = 23\), side \(a\) is opposite angle \(A\), side \(b\) is opposite angle \(B\), and side \(a\) is adjacent to angle \(B\), side \(b\) is adjacent to angle \(A\).

Wait, using cosine for angle \(B = 46^\circ\): \(\cos(46^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{23}\)? Wait, no, adjacent to \(46^\circ\) is \(a\)? Wait, no, if angle \(B\) is \(46^\circ\), then the sides: side \(a\) is adjacent to \(B\) (since \(B\) is at left, right angle at \(C\), so side \(a\) is between \(B\) and \(C\), so adjacent to \(B\)), side \(b\) is between \(A\) and \(C\), adjacent to \(A\), and hypotenuse \(c = 23\) is between \(A\) and \(B\).

So \(\cos(B) = \frac{\text{adjacent to } B}{\text{hypotenuse}} = \frac{a}{23}\), so \(a = 23 \times \cos(46^\circ)\).

\(\sin(B) = \frac{\text{opposite to } B}{\text{hypotenuse}} = \frac{b}{23}\), so \(b = 23 \times \sin(46^\circ)\).

Wait, let's check:

First, angle \(A = 90^\circ - 46^\circ = 44^\circ\) (since right triangle, angles sum to \(180^\circ\), so \(A = 180 - 90 - 46 = 44^\circ\), which matches Step1.

Now, hypotenuse \(c = 23\).

To find \(a\) (adjacent to \(46^\circ\), so \(\cos(46^\circ) = \frac{a}{c}\) => \(a = c \times \cos(46^\circ)\). \(\cos(46^\circ) \approx 0.6947\), so \(a = 23 \times 0.6947 \approx 23 \times 0.6947 \approx 15.9781 \approx 16.0\) (rounded to nearest tenth).

To find \(b\) (opposite to \(46^\circ\), so \(\sin(46^\circ) = \frac{b}{c}\) => \(b = c \times \sin(46^\circ)\). \(\sin(46^\circ) \approx 0.7193\), so \(b = 23 \times 0.7193 \approx 16.5439 \approx 16.5\) (rounded to nearest tenth). Wait, but let's verify with angle \(A = 44^\circ\), so \(\sin(44^\circ) = \frac{b}{c}\)? No, wait, no: if \(a\) is adjacent to \(46^\circ\), then \(a\) is opposite to \(A = 44^\circ\), so \(\sin(44^\circ) = \frac{a}{c}\), let's check: \(\sin(44^\circ) \approx 0.6947\), so \(a = 23 \times 0.6947 \approx 15.978 \approx 16.0\), which matches. And \(\cos(44^\circ) = \frac{b}{c}\), \(\cos(44^\circ) \approx 0.7193\), so \(b = 23 \times 0.7193 \approx 16.543 \approx 16.5\). So that's consistent.

So:

Angle \(A = 44^\circ\)

Side \(a\): using \(\cos(46^\circ) = \frac{a}{23}\) => \(a = 23 \times \cos(46^\circ) \approx 23 \times 0.6947 \approx 15.978 \approx 16.0\)

Side \(b\): using \(\sin(46^\circ) = \frac{b}{23}\) => \(b = 23 \times \sin(46^\circ) \approx 23 \times 0.7193 \approx 16.543 \approx 16.5\)

Wait, but let's confirm the labels again. The triangle has:

- Right angle at the bottom (between \(a\) and \(b\))

- Angle at left: \(46^\circ\) (between side \(a\) and the hypotenuse \(c = 23\))

- Hypotenuse \(c = 23\) (top side, between left angle and angle \(A\))

So:

- \(a\) is the side adjacent to the \(46^\circ\) angle (since it's between the \(46^\circ\) angle and the right angle)

- \(b\) is the side opposite to the \(46^\circ\) angle (between the right angle and angle \(A\))

Therefore:

\(\cos(46^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{23}\) => \(a = 23 \times \cos(46^\circ) \approx 23 \times 0.6947 \approx 16.0\)

\(\sin(46^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{23}\) => \(b = 23 \times \sin(46^\circ) \approx 23 \times 0.7193 \approx 16.5\)

Angle \(A = 90^\circ - 46^\circ = 44^\circ\)

Answer:

\(A = 44^\circ\), \(a \approx 16.0\), \(b \approx 16.5\)

So filling in:

\(A = \boxed{44}\)°

\(a = \boxed{16.0}\)

\(b = \boxed{16.5}\)