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Chapter 4: Problem 5

Find a number \(t\) such that the line through the origin that contains the point \((4, t)\) makes a \(22^{\circ}\) angle with the positive horizontal axis.

Short Answer

Expert verified
The value of \(t\) is approximately \(1.616\), making the point \((4, t)\) as \((4, 1.616)\).

Step by step solution

01

Find the tangent of the angle in terms of the coordinates

Since the line passes through the origin, we can consider that the point \((4, t)\) forms a right triangle with the origin (0,0) as the opposite vertex. The angle between the positive horizontal axis and the hypotenuse is \(22^{\circ}\). Using trigonometry, we can write the tangent of the angle \(22^{\circ}\) as the ratio of the opposite side to the adjacent side: \[\tan(22^{\circ}) = \frac{t}{4}\]
02

Solve for \(t\)

Now, we need to solve for the unknown coordinate \(t\). First, we can find the value of \(\tan(22^{\circ})\) using a calculator or a trigonometric table, and then solve for \(t\): \[\tan(22^{\circ}) \approx 0.4040\] Substitute this value into the equation we derived in step 1: \[0.4040 = \frac{t}{4}\] To find \(t\), we can multiply both sides by 4: \[t = 4 \times 0.4040\] \[t \approx 1.616\] So, the value of \(t\) is approximately \(1.616\). Thus, the point \((4, t)\) is \((4, 1.616)\) and the line through the origin containing this point makes a \(22^{\circ}\) angle with the positive horizontal axis.

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Most popular questions from this chapter

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\tan \frac{\pi}{12}\)

Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with $$\tan u=-2 \text { and } \tan v=-3$$ Find exact expressions for the indicated quantities. \(\cos (v-6 \pi)\)

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Find the smallest positive number \(x\) such that $$ \sin (x+\pi)-\sin x=1 $$

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\tan \left(-\frac{\pi}{8}\right)\)
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