Question

solve for x. round to the nearest tenth, if necessary.
triangle with right angle at b, side bc = 44, angle at c is 46°, side ac is x

Explanation:

Step1: Identify trigonometric ratio

In right triangle \(ABC\) (right-angled at \(B\)), we can use the tangent function. The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. For angle \(C = 46^\circ\), the opposite side to \(C\) is \(AB\) (wait, no, wait: angle \(C\) has adjacent side \(BC = 44\) and opposite side \(AB\)? Wait, no, the side we need to find is \(x = AC\)? Wait, no, wait the side labeled \(x\) is \(AC\)? Wait, no, looking at the triangle: right angle at \(B\), so sides: \(BC = 44\), angle at \(C\) is \(46^\circ\), and side \(x\) is \(AB\)? Wait, no, maybe I mislabel. Wait, in right triangle \(ABC\), right-angled at \(B\), so:
- Angle at \(C\): \(46^\circ\)
- Side \(BC = 44\) (adjacent to angle \(C\))
- Side \(AB\): opposite to angle \(C\)
- Side \(AC\): hypotenuse? Wait, no, wait the side labeled \(x\) is \(AC\)? Wait, no, the problem says "solve for \(x\)" and \(x\) is on side \(AC\)? Wait, no, maybe I got the sides wrong. Wait, let's re-express:

In right triangle \(ABC\), \(\angle B = 90^\circ\), \(\angle C = 46^\circ\), \(BC = 44\), and we need to find \(x\) (which is \(AB\) or \(AC\)? Wait, the side with \(x\) is from \(A\) to \(C\)? No, looking at the diagram: point \(A\) at top, \(B\) at left (right angle), \(C\) at bottom. So side \(BC\) is 44, angle at \(C\) is \(46^\circ\), and side \(x\) is \(AB\)? Wait, no, maybe tangent: \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\). For angle \(C = 46^\circ\), opposite side is \(AB\), adjacent side is \(BC = 44\). Wait, but maybe we need to use cotangent or tangent. Wait, no, wait the side \(x\) is \(AC\)? Wait, no, the label \(x\) is on the side from \(A\) to \(C\)? Wait, no, the diagram: \(A\) connected to \(B\) and \(C\), \(B\) connected to \(C\) (length 44), right angle at \(B\), angle at \(C\) is \(46^\circ\), and \(x\) is the length of \(AB\)? Wait, no, maybe I made a mistake. Wait, let's use trigonometric ratios.

Wait, in right triangle \(ABC\), \(\angle C = 46^\circ\), \(BC = 44\) (adjacent to \(\angle C\)), and we need to find \(AB\) (opposite to \(\angle C\)) or \(AC\) (hypotenuse)? Wait, the side labeled \(x\) is from \(A\) to \(C\)? No, the problem's diagram: \(A\) is top, \(B\) is left (right angle), \(C\) is bottom. So \(AB\) is vertical, \(BC\) is horizontal (length 44), angle at \(C\) is \(46^\circ\), so \(\tan(46^\circ) = \frac{AB}{BC}\), so \(AB = BC \cdot \tan(46^\circ)\). Wait, but maybe \(x\) is \(AB\)? Wait, no, the side with \(x\) is from \(A\) to \(C\)? Wait, maybe I misread. Wait, the problem says "solve for \(x\)" and \(x\) is the length of \(AB\)? Wait, no, let's check again.

Wait, in right triangle \(ABC\), right-angled at \(B\):

- \(\angle C = 46^\circ\)
- \(BC = 44\) (adjacent side to \(\angle C\))
- \(AB\): opposite side to \(\angle C\)
- \(AC\): hypotenuse

But the side labeled \(x\) is \(AB\)? Wait, the diagram shows \(x\) on the side from \(A\) to \(C\)? No, maybe the side with \(x\) is \(AB\). Wait, perhaps I should use the tangent function. So \(\tan(\angle C) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC}\). So \(AB = BC \cdot \tan(46^\circ)\). Wait, but \(BC = 44\), so \(AB = 44 \cdot \tan(46^\circ)\). Let's calculate \(\tan(46^\circ) \approx 1.0355\). So \(AB \approx 44 \times 1.0355 \approx 45.6\)? Wait, no, maybe I got the sides wrong. Wait, maybe \(x\) is the hypotenuse? Wait, no, the hypotenuse is \(AC\). Wait, let's use cosine: \(\cos(\angle C) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AC}\), so \(AC = \frac{BC}{\cos(46^\circ)}\). Let's check: \(\cos(46^\circ) \approx 0.6947\), so \(AC = \frac{44}{0.6947} \approx 63.3\). But that doesn't seem right. Wait, maybe I mixed up opposite and adjacent. Wait, angle at \(C\): the sides:

- Adjacent: \(BC\) (length 44)
- Opposite: \(AB\)
- Hypotenuse: \(AC\)

Wait, the problem's \(x\) is \(AB\)? Then \(\tan(46^\circ) = \frac{AB}{BC}\), so \(AB = BC \cdot \tan(46^\circ)\). Let's compute:

\(\tan(46^\circ) \approx 1.0355\)

\(AB = 44 \times 1.0355 \approx 45.6\)

But wait, maybe the side \(x\) is \(AC\)? Wait, no, the diagram: \(x\) is on the side from \(A\) to \(C\)? Wait, maybe I made a mistake. Wait, let's re-express the triangle:

Right-angled at \(B\), so:

- \(\angle A + \angle C = 90^\circ\), so \(\angle A = 90^\circ - 46^\circ = 44^\circ\)

But maybe using tangent for angle \(A\): \(\tan(44^\circ) = \frac{BC}{AB}\), so \(AB = \frac{BC}{\tan(44^\circ)}\). \(\tan(44^\circ) \approx 0.9657\), so \(AB = \frac{44}{0.9657} \approx 45.6\), which matches the previous result. So maybe \(x\) is \(AB\), and the calculation is \(x = 44 \cdot \tan(46^\circ) \approx 44 \times 1.0355 \approx 45.6\). Wait, but let's confirm the trigonometric ratio.

Alternatively, maybe the side \(x\) is \(AC\), the hypotenuse. Then using cosine: \(\cos(46^\circ) = \frac{BC}{AC}\), so \(AC = \frac{BC}{\cos(46^\circ)} = \frac{44}{\cos(46^\circ)}\). \(\cos(46^\circ) \approx 0.6947\), so \(AC \approx \frac{44}{0.6947} \approx 63.3\). But which is it? Wait, the diagram: the side with \(x\) is from \(A\) to \(C\)? Wait, the problem's diagram: \(A\) at top, \(B\) at left (right angle), \(C\) at bottom. So \(AB\) is vertical (length \(x\)), \(BC\) is horizontal (length 44), angle at \(C\) is \(46^\circ\). So in that case, \(\tan(46^\circ) = \frac{AB}{BC}\), so \(AB = BC \cdot \tan(46^\circ)\). So \(x = 44 \cdot \tan(46^\circ)\).

Calculating \(\tan(46^\circ) \approx 1.035530314\)

So \(x = 44 \times 1.035530314 \approx 45.6\) (rounded to the nearest tenth).

Step2: Perform the calculation

Calculate \(44 \times \tan(46^\circ)\). Using a calculator, \(\tan(46^\circ) \approx 1.0355\), so \(44 \times 1.0355 = 45.562\), which rounds to \(45.6\) (nearest tenth).

Answer:

\(45.6\)