Subjects geometry

Lines Angles 1B8Dbf

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Question: CHAPTER 2 1 Geometry Lines and Angles DIRECTIONS: Find the value of each expression in degrees. For A through E, use the following figure: A. what is the measure of \(\angle a\) ? B. what is the measure of \(b\) ? C. what is the measure of \(c + d\) ? D. what is the measure of \(e\) ? E. What is the measure of \(2c + 3f\) ? position_hint: top-right A line labeled \(p\) slopes down from left to right and intersects the top horizontal line at an angle marked \(40^\circ\); nearby are angles \(a\) and \(b\) on the first and second horizontal lines, with \(c\) on the right-side diagonal intersection on the second horizontal line, and a small right-angle marker at the lower central intersection where lines \(m\), \(n\), and the diagonal lines meet; another diagonal labeled \(q\) intersects the top line with angles \(f\) and \(g\), and the bottom cluster has angles \(d\) and \(e\) around the intersection.
1. **State the problem:** We need to find the measures of angles \(a\), \(b\), \(c + d\), \(e\), and \(2c + 3f\) in degrees based on the given figure with lines and angles. 2. **Given information:** - Angle at the intersection of line \(p\) with the top horizontal line is \(40^\circ\). - There is a right angle (\(90^\circ\)) at the lower central intersection. 3. **Find \(a\):** - Since \(p\) intersects the top horizontal line at \(40^\circ\), and \(a\) is the adjacent angle on the straight line, angles \(a\) and \(40^\circ\) are supplementary. - Use the supplementary angle rule: $$a + 40 = 180$$ - Solve for \(a\): $$a = 180 - 40 = 140$$ 4. **Find \(b\):** - \(a\) and \(b\) are corresponding angles formed by parallel lines and a transversal, so \(b = a = 140\). 5. **Find \(c + d\):** - The right angle marker indicates \(c + d = 90^\circ\). 6. **Find \(e\):** - \(e\) and \(d\) are complementary to the right angle, so \(e = 90 - d\). - Since \(c + d = 90\), then \(d = 90 - c\). - Substitute \(d\) into \(e = 90 - d\): $$e = 90 - (90 - c) = c$$ 7. **Find \(2c + 3f\):** - \(f\) is vertically opposite to the \(40^\circ\) angle, so \(f = 40\). - Calculate: $$2c + 3f = 2c + 3 \times 40 = 2c + 120$$ **Final answers:** - \(a = 140^\circ\) - \(b = 140^\circ\) - \(c + d = 90^\circ\) - \(e = c\) - \(2c + 3f = 2c + 120\)
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