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Chapter 14: Problem 1233

The angle between the lines \(x \cos 85^{\circ}+y \sin 85^{\circ}=1\) and \(x \cos 40^{\circ}+y \sin 40^{\circ}=2\) is : (a) \(90^{\circ}\) (b) \(80^{\circ}\) (c) \(125^{\circ}\) (d) \(45^{\circ}\)

Short Answer

Expert verified
The angle between the lines is (b) \(80^{\circ}\).

Step by step solution

01

Find Direction Ratios of Line 1

Let \(A_1(x_1, y_1)\) be a point on Line 1 such that \(x_1 \cos 85^{\circ} + y_1 \sin 85^{\circ} = 1\). We can find the direction ratios of Line 1 by considering the vector \(\vec{A_1}\).
02

Find Direction Ratios of Line 2

Let \(A_2(x_2, y_2)\) be a point on Line 2 such that \(x_2 \cos 40^{\circ} + y_2 \sin 40^{\circ} = 2\). Similarly, we can find the direction ratios of Line 2 by considering the vector \(\vec{A_2}\).
03

Calculate the Dot Product and Angle between Direction Ratios

Now that we have the direction of vectors for both lines, we can find the angle between them. Let \(\vec{v_1}\) and \(\vec{v_2}\) be the direction vectors for Line 1 and Line 2 respectively. The angle between these two lines, \(\theta\), can be found using the dot product formula: \[\cos \theta = \frac{\vec{v_1} \cdot \vec{v_2}}{||\vec{v_1}|| \; ||\vec{v_2}||}\] Where \(\vec{v_1} \cdot \vec{v_2}\) is the dot product of the direction vectors and ||\(\vec{v_1}\)|| and ||\(\vec{v_2}\)|| are the magnitudes of the direction vectors.
04

Calculate the Values and Choose the Correct Answer

After finding the value of the angle \(\theta\), we can compare it to the given options and choose the one that matches. (a) \(90^{\circ}\) (b) \(80^{\circ}\) (c) \(125^{\circ}\) (d) \(45^{\circ}\)

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Most popular questions from this chapter

Three straight lines \(2 \mathrm{x}+11 \mathrm{y}-5=0,4 \mathrm{x}-3 \mathrm{y}-2=0\) and \(24 x+7 y-20=0\) (a) form a triangle (b) are only concurrent (c) are concurrent with one line bisecting the angle between the other two. (d) none of these

A straight line passes through a point \(\mathrm{A}(1,2)\) and makes an angle \(60^{\circ}\) with the \(\mathrm{x}\) -axis. This line intersect the line \(\mathrm{x}+\mathrm{y}=6\) at \(\mathrm{P}\). Then AP will be (a) \(3(\sqrt{3}+1)\) (b) \(3(\sqrt{3}-1)\) (c) \((\sqrt{3}+1)\) (d) \(3 \sqrt{3}\)

The nearest point on the line \(3 \mathrm{x}+4 \mathrm{y}=1\) from origin is (a) \([(7 / 25),(4 / 25)]\) (b) \([(7 / 25),(2 / 25)]\) (c) \([(3 / 25),(4 / 25)]\) (d) \([(1 / 25),(3 / 25)]\)

\(\mathrm{A}(1,0)\) and \(\mathrm{B}(-1,0)\), then the locus of points satisfying \(\mathrm{AQ}-\mathrm{BQ}=\pm 1\) is \(\ldots \ldots\) (a) \(12 \mathrm{x}^{2}+4 \mathrm{y}^{2}=3\) (b) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=3\) (c) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=-3\) (d) \(12 x^{2}+4 y^{2}=-3\)

If \(\mathrm{P}(-1,0), \mathrm{Q}(0,0)\) and \(\mathrm{R}(3,3 \sqrt{3})\), then the equation of bisector of \(\angle P Q R\) is \(\ldots \ldots .\) (a) \((\sqrt{3} / 2) \mathrm{x}+\mathrm{y}=0\) (b) \(x+(\sqrt{3} / 2) y=0\) (c) \(\sqrt{3 x+y}=0\) (d) \(x+\sqrt{3} y=0\)
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