Math Problem Statement

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Question 2: Completions

(a) Simplify the trigonometric expression:

sin30+2cos60tan245\sin 30^\circ + 2\cos 60^\circ - \tan^2 45^\circ Solution:

  • sin30=12\sin 30^\circ = \frac{1}{2},
  • cos60=12\cos 60^\circ = \frac{1}{2},
  • tan45=1\tan 45^\circ = 1.

Substitute these values: sin30+2cos60tan245=12+21212\sin 30^\circ + 2\cos 60^\circ - \tan^2 45^\circ = \frac{1}{2} + 2 \cdot \frac{1}{2} - 1^2 =12+11=12.= \frac{1}{2} + 1 - 1 = \frac{1}{2}.

Final Answer: 12\frac{1}{2}.


(b) Find CC, the midpoint of ABAB, where A(1,5)A(1, 5) and B(3,7)B(3, 7).

Formula for the midpoint: C(x,y)=(x1+x22,y1+y22).C(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

Substitute A(1,5)A(1, 5) and B(3,7)B(3, 7): C=(1+32,5+72)=(42,122)=(2,6).C = \left(\frac{1 + 3}{2}, \frac{5 + 7}{2}\right) = \left(\frac{4}{2}, \frac{12}{2}\right) = (2, 6).

Final Answer: C(2,6)C(2, 6).


(c) Find the slope of ABAB, where A(2,3)A(2, -3) and B(1,5)B(1, -5):

Formula for slope: m=y2y1x2x1.m = \frac{y_2 - y_1}{x_2 - x_1}.

Substitute A(2,3)A(2, -3) and B(1,5)B(1, -5): m=5(3)12=5+312=21=2.m = \frac{-5 - (-3)}{1 - 2} = \frac{-5 + 3}{1 - 2} = \frac{-2}{-1} = 2.

Final Answer: Slope m=2m = 2.


(d) Solve for cosCtanA\cos C \cdot \tan A in triangle ABCABC, where:

  • Right triangle, B=90\angle B = 90^\circ,
  • AB=5AB = 5, BC=12BC = 12.

Find ACAC using the Pythagorean theorem: AC=AB2+BC2=52+122=25+144=169=13.AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

cosC=BCAC=1213\cos C = \frac{BC}{AC} = \frac{12}{13}, and tanA=BCAB=125\tan A = \frac{BC}{AB} = \frac{12}{5}.

cosCtanA=1213125=14465\cos C \cdot \tan A = \frac{12}{13} \cdot \frac{12}{5} = \frac{144}{65}.

Final Answer: 14465\frac{144}{65}.


(e) Solve for mm, where tan(x+25)=1\tan(x + 25^\circ) = 1 and x+25x + 25^\circ is acute:

If tan(x+25)=1\tan(x + 25^\circ) = 1, then: x+25=45.x + 25^\circ = 45^\circ.

Solve for xx: x=4525=20.x = 45^\circ - 25^\circ = 20^\circ.

Final Answer: x=20x = 20^\circ.


(f) Find the coordinates of the center of a circle where ABAB is its diameter:

  • A(1,4)A(-1, -4) and B(7,6)B(7, 6).

The center is the midpoint of ABAB: Midpoint=(1+72,4+62)=(62,22)=(3,1).\text{Midpoint} = \left(\frac{-1 + 7}{2}, \frac{-4 + 6}{2}\right) = \left(\frac{6}{2}, \frac{2}{2}\right) = (3, 1).

Final Answer: Center at (3,1)(3, 1).


Do you want me to continue solving the next questions or provide more detailed explanations for any part?

Here are 5 related questions:

  1. How do you derive the midpoint formula geometrically?
  2. Why does tan(x)=1\tan(x) = 1 imply specific angle values?
  3. Can you calculate the slope if x1=x2x_1 = x_2? What happens?
  4. How do you confirm if three points form a right triangle?
  5. How does the Pythagorean theorem extend to coordinate geometry?

Tip: Always double-check trigonometric simplifications for precision in larger problems!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Coordinate Geometry
Midpoint and Slope Calculation
Pythagorean Theorem
Equations of Lines

Formulas

Midpoint formula: (x, y) = ((x1 + x2)/2, (y1 + y2)/2)
Slope formula: m = (y2 - y1)/(x2 - x1)
Pythagorean theorem: c^2 = a^2 + b^2
Trigonometric simplifications (sin, cos, tan)

Theorems

Pythagorean theorem
Properties of a right triangle
Basic trigonometric identities

Suitable Grade Level

Grades 9-12