Question

find the length ( x ) to the nearest whole number.
(do not round until the final answer. then round to the nearest whole number.)

Explanation:

Step1: Analyze the triangle structure

The diagram shows a triangle with a common side (let's call it \( d = 860 \)) and two angles \( 35^\circ \) and \( 28^\circ \) at the vertex. The length \( x \) can be related to the height of two right - angled triangles formed. If we consider the two right - angled triangles, the height of the first right - angled triangle (with angle \( 35^\circ \)) is \( \frac{x}{2}\tan(35^\circ) \) and the height of the second right - angled triangle (with angle \( 28^\circ \)) is \( \frac{x}{2}\tan(28^\circ) \)? Wait, no, actually, if we assume that the total length from the vertex to the base is split into two equal parts? No, maybe a better approach: Let's assume that the distance from the vertex to the line of length \( x \) is \( d = 860 \). Then, if we have two angles \( 35^\circ \) and \( 28^\circ \), and the length \( x \) is such that \( \frac{x}{2}=d\tan(35^\circ)-d\tan(28^\circ) \)? Wait, no, let's re - think. Suppose we have a point, and from that point, we draw two lines making angles \( 35^\circ \) and \( 28^\circ \) with a vertical line (or horizontal line). Let's assume that the distance from the point to the base (the line containing \( x \)) is \( 860 \). Then, the length of the segment corresponding to the \( 35^\circ \) angle is \( 860\tan(35^\circ) \) and the length corresponding to the \( 28^\circ \) angle is \( 860\tan(28^\circ) \). Then, the length \( x \) is the difference between these two lengths multiplied by 2? Wait, no, if the two angles are on either side of the perpendicular from the point to the base, then \( x = 2\times(860\tan(35^\circ)-860\tan(28^\circ)) \).

Step2: Calculate \( \tan(35^\circ) \) and \( \tan(28^\circ) \)

We know that \( \tan(35^\circ)\approx0.7002 \) and \( \tan(28^\circ)\approx0.5317 \)

Step3: Calculate the difference

First, calculate \( 860\tan(35^\circ)-860\tan(28^\circ)=860\times(0.7002 - 0.5317)=860\times0.1685 = 144.91 \)

Step4: Calculate \( x \)

Since \( x = 2\times(860\tan(35^\circ)-860\tan(28^\circ)) \) (assuming the symmetry, that the length \( x \) is twice the difference of the two tangent - based lengths), then \( x = 2\times144.91=289.82\approx290 \)? Wait, no, maybe my initial assumption about the geometry is wrong. Wait, another approach: Let's assume that we have a triangle where the adjacent side to the angles is \( 860 \), and the length \( x \) is the opposite side for the difference of the two angles. Wait, if we consider the formula for the length of a side in a triangle with two angles and a side. Wait, maybe the correct formula is \( x = 860\times(\tan(35^\circ)+\tan(28^\circ)) \)? No, let's use the tangent function correctly.

Let's consider two right - angled triangles. In the first right - angled triangle, angle \( \theta_1 = 35^\circ \), adjacent side \( a = 860 \), opposite side \( y_1 \). So \( \tan(35^\circ)=\frac{y_1}{860}\Rightarrow y_1 = 860\tan(35^\circ) \)

In the second right - angled triangle, angle \( \theta_2 = 28^\circ \), adjacent side \( a = 860 \), opposite side \( y_2 \). So \( \tan(28^\circ)=\frac{y_2}{860}\Rightarrow y_2 = 860\tan(28^\circ) \)

If the length \( x \) is the sum of \( y_1 \) and \( y_2 \) (if the two triangles are on opposite sides of the perpendicular), then \( x=y_1 + y_2=860\tan(35^\circ)+860\tan(28^\circ) \)

Calculate \( \tan(35^\circ)\approx0.7002 \), \( \tan(28^\circ)\approx0.5317 \)

\( y_1 = 860\times0.7002 = 860\times0.7002 = 602.172 \)

\( y_2 = 860\times0.5317 = 860\times0.5317 = 457.262 \)

\( x=602.172 + 457.262=1059.434\approx1059 \)? Wait, this is different from the previous result. I think I mis - interpreted the diagram.

Wait, the correct diagram: It's a triangle with a vertex angle split into \( 35^\circ \) and \( 28^\circ \), and the distance from the vertex to the base (the side of length \( x \)) is \( 860 \). So the total angle at the vertex is \( 35^\circ+28^\circ = 63^\circ \)? No, maybe the two angles are on the same side? Wait, no, the diagram shows a triangle with a line segment of length \( x \) and a distance of \( 860 \) from the vertex to the line segment. So, if we have two angles, \( 35^\circ \) and \( 28^\circ \), and the length \( x \) is given by \( x = 860\times(\tan(35^\circ)-\tan(28^\circ))\times2 \)? Wait, let's use the correct formula for the length of a chord or the difference in tangents.

Wait, let's assume that we have a point \( O \), and a line \( AB \) with length \( x \), and the distance from \( O \) to \( AB \) is \( 860 \). Let \( M \) be the mid - point of \( AB \), so \( OM = 860 \), and \( \angle AOM = 35^\circ \), \( \angle BOM = 28^\circ \). Then, \( AM = OM\tan(35^\circ) \), \( BM = OM\tan(28^\circ) \), and \( x=AM - BM \)? No, that would be if the angles are on the same side. Wait, if \( \angle AOM = 35^\circ \) and \( \angle BOM = 28^\circ \) with \( A \) and \( B \) on opposite sides of \( M \), then \( AM = OM\tan(35^\circ) \), \( BM = OM\tan(28^\circ) \), and \( x=AM + BM \)? No, that can't be. Wait, I think I made a mistake in the angle interpretation.

Wait, the correct approach: If we have a triangle where the height (distance from vertex to base) is \( h = 860 \), and the base is \( x \), and the vertex angle is split into two angles \( 35^\circ \) and \( 28^\circ \), then the half - base for the first angle is \( \frac{x_1}{2}=h\tan(35^\circ) \) and the half - base for the second angle is \( \frac{x_2}{2}=h\tan(28^\circ) \), and \( x=x_1 - x_2 \)? No, this is getting confusing. Let's use the formula for the length of a side in a triangle with two angles and a height.

Alternatively, the length \( x \) is given by \( x = 860\times(\tan(35^\circ)+\tan(28^\circ)) \)

Calculate \( \tan(35^\circ)\approx0.7002 \), \( \tan(28^\circ)\approx0.5317 \)

\( \tan(35^\circ)+\tan(28^\circ)=0.7002 + 0.5317 = 1.2319 \)

\( x = 860\times1.2319=1059.434\approx1059 \)

Wait, but maybe the correct formula is \( x = 860\times(\tan(35^\circ)-\tan(28^\circ))\times2 \)

\( \tan(35^\circ)-\tan(28^\circ)=0.7002 - 0.5317 = 0.1685 \)

\( 860\times0.1685 = 144.91 \)

\( 144.91\times2 = 289.82\approx290 \)

This is a problem. The key is to correctly interpret the diagram. Since the diagram shows a triangle with a line segment of length \( x \) and a distance of \( 860 \) from the vertex to the line segment, and two angles \( 35^\circ \) and \( 28^\circ \) at the vertex. So, the correct formula is that the length \( x \) is equal to \( 860\times(\tan(35^\circ)+\tan(28^\circ)) \) if the two angles are on opposite sides of the perpendicular, or \( 860\times(\tan(35^\circ)-\tan(28^\circ))\times2 \) if the two angles are on the same side.

Wait, let's check with the tangent values. \( \tan(35^\circ)\approx0.7002 \), \( \tan(28^\circ)\approx0.5317 \)

If we consider that the two angles are on the same side, and the length \( x \) is the difference between the two tangent - based lengths multiplied by 2 (because it's symmetric on both sides of the perpendicular), then:

\( \frac{x}{2}=860\tan(35^\circ)-860\tan(28^\circ) \)

\( x = 2\times860\times(\tan(35^\circ)-\tan(28^\circ)) \)

\( x = 2\times860\times(0.7002 - 0.5317) \)

\( x = 2\times860\times0.1685 \)

\( x = 2\times144.91 \)

\( x = 289.82\approx290 \)

But if we consider the two angles on opposite sides, \( x = 860\times(\tan(35^\circ)+\tan(28^\circ)) \)

\( x = 860\times(0.7002 + 0.5317) \)

\( x = 860\times1.2319 \)

\( x = 1059.434\approx1060 \)

This is a problem. But looking at the diagram, it's more likely that the two angles are on the same side, and the length \( x \) is the difference between the two tangent - based lengths multiplied by 2 (since it's a symmetric figure with respect to the perpendicular). So we will go with \( x = 2\times860\times(\tan(35^\circ)-\tan(28^\circ)) \)

Step1: Calculate \( \tan(35^\circ) \) and \( \tan(28^\circ) \)

\( \tan(35^\circ)\approx0.7002 \), \( \tan(28^\circ)\approx0.5317 \)

Step2: Calculate the difference of tangents

\( \tan(35^\circ)-\tan(28^\circ)=0.7002 - 0.5317 = 0.1685 \)

Step3: Multiply by 2 and 860

\( x = 2\times860\times0.1685 \)

\( x = 1720\times0.1685 \)

\( x = 289.82\approx290 \)

Answer:

\( \boxed{290} \)