angles of triangles word problems
date:
1. in △qpl, m∠q is five times the m∠p and m∠l is twelve more than twice m∠p. find the measure of each angle.
2. the measure of the larger acute angle in a right triangle is 10 degrees less than three times the measure of the smaller acute angle. find the measure of each angle.
equilateral & isosceles triangles
date:
1. if △mvp is an equilateral triangle, find the values of x and y.
(there is a triangle mvp with side mv labeled 14, side mp labeled 3x - 4, and side vp labeled (5y - 5)°)
2. △jkl is an isosceles triangle with vertex angle k. find the values of x and y if jk = 8x, kl = 13x - 15, m∠kjl = (5y - 3)°, and m∠jkl = 76°.

Question

Accepted Answer

### Problem 1 (Angles of Triangles Word Problems - First Question)
# Explanation:
## Step 1: Define Variables
Let $ m\angle P = x $. Then $ m\angle Q = 5x $ and $ m\angle L = 2x + 12 $.
## Step 2: Use Triangle Angle Sum
In a triangle, the sum of angles is $ 180^\circ $. So, $ x + 5x + (2x + 12) = 180 $.
## Step 3: Simplify and Solve
Combine like terms: $ 8x + 12 = 180 $. Subtract 12: $ 8x = 168 $. Divide by 8: $ x = 21 $.
## Step 4: Find Each Angle
$ m\angle P = 21^\circ $, $ m\angle Q = 5\times21 = 105^\circ $, $ m\angle L = 2\times21 + 12 = 54^\circ $.
# Answer:
$ m\angle P = 21^\circ $, $ m\angle Q = 105^\circ $, $ m\angle L = 54^\circ $

### Problem 2 (Angles of Triangles Word Problems - Second Question)
# Explanation:
## Step 1: Define Variables
Let the smaller acute angle be $ x $. Then the larger is $ 3x - 10 $.
## Step 2: Use Right Triangle Angle Sum
In a right triangle, the two acute angles sum to $ 90^\circ $. So, $ x + (3x - 10) = 90 $.
## Step 3: Simplify and Solve
Combine like terms: $ 4x - 10 = 90 $. Add 10: $ 4x = 100 $. Divide by 4: $ x = 25 $.
## Step 4: Find Each Angle
Smaller angle: $ 25^\circ $, Larger angle: $ 3\times25 - 10 = 65^\circ $.
# Answer:
Smaller acute angle: $ 25^\circ $, Larger acute angle: $ 65^\circ $

### Problem 3 (Equilateral & Isosceles Triangles - First Question)
# Explanation:
## Step 1: Properties of Equilateral Triangle
In an equilateral triangle, all sides are equal and all angles are $ 60^\circ $. So, $ 3x - 4 = 14 $ (sides $ MN $ and $ MP $) and $ 5y - 5 = 60 $ (angle at $ N $).
## Step 2: Solve for $ x $
$ 3x - 4 = 14 $. Add 4: $ 3x = 18 $. Divide by 3: $ x = 6 $.
## Step 3: Solve for $ y $
$ 5y - 5 = 60 $. Add 5: $ 5y = 65 $. Divide by 5: $ y = 13 $.
# Answer:
$ x = 6 $, $ y = 13 $

### Problem 4 (Equilateral & Isosceles Triangles - Second Question)
# Explanation:
## Step 1: Properties of Isosceles Triangle
In $ \triangle JKL $ with vertex $ K $, $ JK = KL $ (legs) and base angles $ \angle JKL $ and $ \angle KJL $ are equal? Wait, no: Vertex angle is $ K $, so base angles are $ \angle J $ and $ \angle L $. Wait, given $ JK = 8x $, $ KL = 13x - 15 $, so $ JK = KL $ (isosceles with $ JK = KL $)? Wait, no, vertex angle $ K $, so legs are $ JK $ and $ LK $, base is $ JL $. So $ JK = LK $? Wait, $ JK = 8x $, $ KL = 13x - 15 $, so $ 8x = 13x - 15 $. Also, $ m\angle KJL = (5y - 3)^\circ $, $ m\angle JKL = 76^\circ $. In isosceles triangle, base angles are equal: $ \angle J = \angle L $, and $ \angle J + \angle L + \angle K = 180 $. So $ 2(5y - 3) + 76 = 180 $.
## Step 2: Solve for $ x $
$ 8x = 13x - 15 $. Subtract $ 8x $: $ 0 = 5x - 15 $. Add 15: $ 5x = 15 $. Divide by 5: $ x = 3 $.
## Step 3: Solve for $ y $
$ 2(5y - 3) + 76 = 180 $. Simplify: $ 10y - 6 + 76 = 180 $. $ 10y + 70 = 180 $. Subtract 70: $ 10y = 110 $. Divide by 10: $ y = 11 $.
# Answer:
$ x = 3 $, $ y = 11 $

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QUESTION IMAGE

Question

Explanation:

Problem 1 (Angles of Triangles Word Problems - First Question)

Step 1: Define Variables

Step 2: Use Triangle Angle Sum

Step 3: Simplify and Solve

Step 4: Find Each Angle

Step 1: Define Variables

Step 2: Use Right Triangle Angle Sum

Step 3: Simplify and Solve

Step 4: Find Each Angle

Step 1: Properties of Equilateral Triangle

Step 2: Solve for \( x \)

Step 3: Solve for \( y \)

Answer:

Problem 2 (Angles of Triangles Word Problems - Second Question)