Question

find the lengths of r, s, t, and u shown in the figure below if s + t = 26. round your answers to the nearest tenth. note that the figure is not drawn to scale.

Explanation:

Step1: Analyze the right triangle with angle 29° and hypotenuse 7? Wait, no, the middle segment is 7. Wait, the right triangle has angle 29°, let's denote the right triangle: let's say the vertical side is \( r \), horizontal side is \( s \), and the middle segment (the angle bisector? No, the segment of length 7 is from the right angle vertex? Wait, no, the figure is a right triangle with a segment drawn to the hypotenuse? Wait, no, the right angle is at the bottom right, angle at the top is 29°, the vertical side is \( r \), horizontal side is \( s + t = 26 \) (since \( s + t = 26 \)), and the segment of length 7 is from the top vertex to a point on the horizontal side, creating two triangles: one with hypotenuse \( u \) and base \( t \), and the right triangle with angle 29°, vertical side \( r \), horizontal side \( s \), and the segment of length 7 is maybe the height? Wait, no, let's use trigonometry. Let's consider the right triangle with angle 29°, so \( \cos(29^\circ) = \frac{r}{\text{hypotenuse}} \)? Wait, no, the right triangle: angle at the top is 29°, right angle at the bottom right, so the vertical side is adjacent to 29°, horizontal side is opposite? Wait, no: in a right triangle, for angle \( \theta \), \( \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \), \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \), \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \). Wait, the segment of length 7: maybe it's the height? Wait, no, let's assume that the segment of length 7 is the altitude to the hypotenuse? Wait, no, the hypotenuse of the big triangle would be \( u \), and the horizontal side is \( s + t = 26 \), vertical side is \( r \). Wait, maybe the triangle with angle 29° has hypotenuse 7? No, the segment of length 7 is from the top vertex to a point on the horizontal side, so the two triangles: the right triangle (let's call it Triangle 1) with angle 29°, vertical side \( r \), horizontal side \( s \), and hypotenuse (let's say) \( c_1 \), and the other triangle (Triangle 2) with base \( t \), vertical side \( r \), and hypotenuse \( u \), and the segment of length 7 is the common side? Wait, no, the segment of length 7 is from the top vertex to the point dividing the horizontal side into \( s \) and \( t \), so the length 7 is the length of that segment. So we have two triangles: one with sides \( r \), \( s \), and 7 (right triangle? Wait, the right angle is at the bottom right, so the segment of length 7 is from the top vertex to a point on the horizontal side, so the triangle with sides \( r \), \( s \), and 7: is that a right triangle? Yes, because the right angle is at the bottom right, so the segment of length 7 is the hypotenuse of a right triangle with legs \( r \) and \( s \)? Wait, no, the right angle is at the bottom right, so the vertical side is \( r \) (adjacent to 29°), horizontal side is \( s \) (opposite to 29°), and the hypotenuse would be the segment from top to bottom right? No, the top vertex is connected to the bottom right (right angle) with \( r \), to the bottom left with \( u \), and the segment of length 7 is from top to a point on the horizontal side (between bottom right and bottom left), so the horizontal side is \( s + t = 26 \), bottom right to the point is \( s \), point to bottom left is \( t \), top to bottom right is \( r \), top to point is 7, top to bottom left is \( u \). So we have two right triangles: one with legs \( r \) and \( s \), hypotenuse 7 (wait, no, 7 is the length from top to the point, so that's the hypotenuse of the triangle with legs \( r \) and \( s \))? Wait, yes! So the right triangle with legs \( r \) (vertical) and \( s \) (horizontal), hypotenuse 7, and angle at the top is 29°, so \( \cos(29^\circ) = \frac{r}{7} \) and \( \sin(29^\circ) = \frac{s}{7} \). Let's calculate \( r \) and \( s \) first.

Step2: Calculate \( r \) and \( s \) using the right triangle with hypotenuse 7 and angle 29°.

\( \cos(29^\circ) = \frac{r}{7} \) ⇒ \( r = 7 \cos(29^\circ) \)
\( \sin(29^\circ) = \frac{s}{7} \) ⇒ \( s = 7 \sin(29^\circ) \)

Calculate \( \cos(29^\circ) \approx 0.8746 \), so \( r \approx 7 \times 0.8746 \approx 6.1222 \approx 6.1 \) (rounded to nearest tenth)
\( \sin(29^\circ) \approx 0.4848 \), so \( s \approx 7 \times 0.4848 \approx 3.3936 \approx 3.4 \) (rounded to nearest tenth)

Step3: Calculate \( t \) since \( s + t = 26 \), so \( t = 26 - s \approx 26 - 3.4 = 22.6 \)

Step4: Calculate \( u \) using the right triangle with legs \( r \) and \( t \), so \( u = \sqrt{r^2 + t^2} \)

First, \( r \approx 6.1 \), \( t \approx 22.6 \)
\( r^2 \approx 6.1^2 = 37.21 \)
\( t^2 \approx 22.6^2 = 510.76 \)
\( u = \sqrt{37.21 + 510.76} = \sqrt{547.97} \approx 23.4 \) (rounded to nearest tenth)

Wait, but let's check the calculations more accurately. Let's use more precise values for \( \cos(29^\circ) \) and \( \sin(29^\circ) \).

\( \cos(29^\circ) \approx 0.87462 \), so \( r = 7 \times 0.87462 \approx 6.1223 \)
\( \sin(29^\circ) \approx 0.48481 \), so \( s = 7 \times 0.48481 \approx 3.3937 \)
\( t = 26 - 3.3937 = 22.6063 \)
\( u = \sqrt{(6.1223)^2 + (22.6063)^2} \)
\( 6.1223^2 = 37.482 \)
\( 22.6063^2 = 511.045 \)
Sum: \( 37.482 + 511.045 = 548.527 \)
\( \sqrt{548.527} \approx 23.42 \approx 23.4 \)

Now, let's verify if the segment of length 7 is indeed the hypotenuse of the right triangle with angle 29°. The angle at the top is 29°, so the adjacent side is \( r \), opposite is \( s \), hypotenuse 7. That makes sense.

So summarizing:

- \( r = 7 \cos(29^\circ) \approx 6.1 \)
- \( s = 7 \sin(29^\circ) \approx 3.4 \)
- \( t = 26 - s \approx 22.6 \)
- \( u = \sqrt{r^2 + t^2} \approx 23.4 \)

Answer:

\( r \approx 6.1 \), \( s \approx 3.4 \), \( t \approx 22.6 \), \( u \approx 23.4 \)