Math Problem Statement

solve the question in the image

Solve the questions in the image: find the value of x, describe the triangle, and solve for (tan A - cos A) given sin A = 3/5.

Solution

Let's solve the questions one by one from the image.

Question 31

Problem: Find the value of xx in the diagram.

Steps:

  1. The given figure has a quadrilateral where the angles are given by (2x)o(2x)^o, (x+10)o(x + 10)^o, (4x+50)o(4x + 50)^o, and (3x)o(3x)^o.
  2. In any quadrilateral, the sum of all interior angles is 360360^\circ.

(2x)+(x+10)+(4x+50)+(3x)=360(2x) + (x + 10) + (4x + 50) + (3x) = 360^\circ

2x+x+10+4x+50+3x=3602x + x + 10 + 4x + 50 + 3x = 360

10x+60=36010x + 60 = 360

10x=36060=30010x = 360 - 60 = 300

x=30010=30x = \frac{300}{10} = 30

So, the value of xx is 30\boxed{30}.

Question 32

Problem: The interior angles of a triangle are (y+10)(y + 10)^\circ, (2y40)(2y - 40)^\circ, and (3y90)(3y - 90)^\circ. Which of the following accurately describes the triangle?

Steps:

  1. The sum of the interior angles of a triangle is 180180^\circ.
  2. Set up the equation:

(y+10)+(2y40)+(3y90)=180(y + 10) + (2y - 40) + (3y - 90) = 180^\circ

y+10+2y40+3y90=180y + 10 + 2y - 40 + 3y - 90 = 180

6y120=1806y - 120 = 180

6y=3006y = 300

y=3006=50y = \frac{300}{6} = 50

  1. Substitute y=50y = 50 back into the expressions for the angles:
  • y+10=50+10=60y + 10 = 50 + 10 = 60^\circ
  • 2y40=10040=602y - 40 = 100 - 40 = 60^\circ
  • 3y90=15090=603y - 90 = 150 - 90 = 60^\circ

All angles are 6060^\circ, which indicates that the triangle is an equilateral triangle.

Thus, the correct answer is C\boxed{C} (equilateral triangle).

Question 33

Problem: Given that sinA=35\sin A = \frac{3}{5} and 0A900^\circ \leq A \leq 90^\circ, find the value of (tanAcosA)(\tan A - \cos A).

Steps:

  1. First, find cosA\cos A using the identity:

sin2A+cos2A=1\sin^2 A + \cos^2 A = 1

(35)2+cos2A=1\left(\frac{3}{5}\right)^2 + \cos^2 A = 1

925+cos2A=1\frac{9}{25} + \cos^2 A = 1

cos2A=1925=1625\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}

cosA=45\cos A = \frac{4}{5}

  1. Next, find tanA\tan A:

tanA=sinAcosA=3545=34\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

  1. Now, calculate (tanAcosA)(\tan A - \cos A):

tanAcosA=3445\tan A - \cos A = \frac{3}{4} - \frac{4}{5}

To subtract, find a common denominator (20):

tanAcosA=15201620=120\tan A - \cos A = \frac{15}{20} - \frac{16}{20} = \frac{-1}{20}

Thus, the correct answer is D\boxed{D} 120\frac{-1}{20}.


Do you need further explanations or have any questions?

Here are five related questions you might find useful:

  1. How can you determine the angles in a cyclic quadrilateral?
  2. What is the relationship between the interior and exterior angles of polygons?
  3. How would you solve for yy if given different expressions for the triangle's angles?
  4. How can you verify trigonometric identities involving sine and cosine?
  5. What other trigonometric functions can be derived using sine and cosine?

Tip: When dealing with problems involving triangles, always remember to use the sum of angles rule to set up your equations.

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Math Problem Analysis

Mathematical Concepts

Geometry
Trigonometry
Algebra

Formulas

Sum of angles in a quadrilateral = 360°
Sum of angles in a triangle = 180°
Pythagorean identity: sin²A + cos²A = 1
tanA = sinA / cosA

Theorems

Sum of interior angles in polygons
Basic trigonometric identities

Suitable Grade Level

Grades 9-10