QUESTION IMAGE
Question
find the length x to the nearest whole number. 250 64° 39° x
Step1: Set up tangent - ratio equations
Let the height from the top of the 250 - unit length to the top - most right - angled vertex be \(h\). For the smaller right - triangle with angle \(64^{\circ}\), \(\tan64^{\circ}=\frac{h}{250}\), so \(h = 250\tan64^{\circ}\). For the larger right - triangle with angle \(39^{\circ}\), \(\tan39^{\circ}=\frac{h + x}{250}\), so \(h+x=250\tan39^{\circ}\).
Step2: Substitute \(h\) into the second equation
Substitute \(h = 250\tan64^{\circ}\) into \(h+x=250\tan39^{\circ}\). We get \(250\tan64^{\circ}+x = 250\tan39^{\circ}\). Then \(x=250\tan39^{\circ}-250\tan64^{\circ}\).
We know that \(\tan64^{\circ}\approx2.0503\) and \(\tan39^{\circ}\approx0.8098\).
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Since length cannot be negative, we may have set up the equations wrong. Let's start over.
Let the total height be \(H\). For the triangle with \(39^{\circ}\), \(\tan39^{\circ}=\frac{H}{250}\), so \(H = 250\tan39^{\circ}\). For the triangle with \(64^{\circ}\), \(\tan64^{\circ}=\frac{H - x}{250}\), so \(H - x=250\tan64^{\circ}\).
Substitute \(H = 250\tan39^{\circ}\) into \(H - x=250\tan64^{\circ}\), we get \(250\tan39^{\circ}-x = 250\tan64^{\circ}\), then \(x=250\tan39^{\circ}-250\tan64^{\circ}\) (wrong).
The correct way: Let the height of the part above \(x\) be \(y\).
For the triangle with \(64^{\circ}\), \(\tan64^{\circ}=\frac{y}{250}\), so \(y = 250\tan64^{\circ}\).
For the triangle with \(39^{\circ}\), \(\tan39^{\circ}=\frac{y + x}{250}\), so \(y+x=250\tan39^{\circ}\).
Substitute \(y = 250\tan64^{\circ}\) into \(y + x=250\tan39^{\circ}\), we get \(x=250\tan39^{\circ}-250\tan64^{\circ}\) (wrong).
The right - hand side should be \(x = 250\tan64^{\circ}-250\tan39^{\circ}\).
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