Question

directions: find the value of each variable.
5.

4.

7.

8.

9.

10.

1. if ( \triangle abc ) is an isosceles triangle and ( \triangle dbe ) is an equilateral triangle, find each angle measure.

Explanation:

Problem 5

Step1: Equal sides, set expressions equal

$4x + 23 = 10x - 1$

Step2: Solve for x

$23 + 1 = 10x - 4x$
$24 = 6x$
$x = \frac{24}{6} = 4$

Problem 6

Step1: Equal sides → base angle = $10x^\circ$

Step2: Triangle angle sum = $180^\circ$

$10x + 10x + (9x - 25) = 180$

Step3: Simplify and solve for x

$29x - 25 = 180$
$29x = 205$
$x = \frac{205}{29} \approx 7.07$

Problem 7

Step1: Equal sides → opposite angles equal

$5x - 7 = 8x - 55$

Step2: Solve for x

$55 - 7 = 8x - 5x$
$48 = 3x$
$x = 16$

Problem 8

Step1: Equal sides → triangle is isosceles, vertex angle = $(4x+8)^\circ$, base angles equal. Let base angles = $y$, so $2y + (4x+8) = 180$. But since sides are marked equal, the vertex angle is the marked one, so base angles are equal, and we can infer: the triangle has two equal sides, so the angles opposite are equal. Wait, the marked angle is the vertex, so the other two angles are equal. But we only have the vertex angle given, so we assume we solve for x such that the triangle is valid? No, wait: the two sides with ticks are equal, so the angles opposite are equal. The marked angle is $(4x+8)^\circ$, which is the angle between the two equal sides (vertex angle). So the sum is $2y + (4x+8) = 180$, but we need another relation? No, wait, no—wait, no, actually, no, the problem is that the two sides with ticks are equal, so the angles opposite are equal. The angle marked is the vertex angle, so we can only solve if we assume that the triangle's angles sum to 180, but we need to realize that this is an isosceles triangle, so the vertex angle is $(4x+8)^\circ$, and the base angles are equal. But since there's no other angle given, wait no—wait, no, I misread: the two sides with ticks are equal, so the angles opposite are equal. The angle marked is the angle between the non-equal side and one equal side? No, no, the tick marks are on two sides, so those two sides are equal. The angle marked is the angle adjacent to one ticked side and the non-ticked side. So the angle opposite the non-ticked side is $(4x+8)^\circ$, and the two equal angles are the ones opposite the ticked sides. Wait, no, let's correct:

Step1: Equal sides → opposite angles equal

Let the two equal angles be $A$, so $2A + (4x+8) = 180$. But no, wait, no—the tick marks are on two sides, so those sides are congruent, so their opposite angles are congruent. The marked angle is $(4x+8)^\circ$, which is the angle between the two ticked sides (vertex angle). So:

Step2: Triangle angle sum

$2A + (4x+8) = 180$, but we need to realize that this is not solvable unless we know that the triangle is valid, but wait no—wait, no, I made a mistake. The problem says "find the value of each variable", so x is the variable, so we must have that the angle is valid, but no—wait, no, actually, no: the two sides with ticks are equal, so the angles opposite are equal. The marked angle is the vertex angle, so the base angles are equal, but we don't have their measures. Wait, no, no, I misread the diagram: the tick marks are on two sides, so those sides are equal, so the angles opposite are equal. The marked angle is the angle opposite the non-ticked side, so that angle is $(4x+8)^\circ$, and the two equal angles are the ones opposite the ticked sides. But since we have no other information, wait no—wait, no, this is a right triangle? The marked angle has a square, so it's a right angle! Oh right! The square means it's a 90° angle.

Step1: Recognize right angle: $4x+8=90$

Step2: Solve for x

$4x = 90 - 8 = 82$
$x = \frac{82}{4} = 20.5$

Problem 9

Step1: Two angles are 60°, so triangle is equilateral, so all sides equal

$7x + 19 = 11x - 89$

Step2: Solve for x

$19 + 89 = 11x - 7x$
$108 = 4x$
$x = 27$

Problem 10

Step1: Equal sides → opposite angles equal. The angle $(11x-65)^\circ$ is supplementary to the base angle? No, wait: the side with the tick is equal to the other side, so the angle opposite the ticked side is 29°, so the angle adjacent to $(11x-65)^\circ$ is 29°, so $(11x-65) + 29 = 180$? No, wait no: the tick marks mean the two sides are equal, so the triangle is isosceles with the two sides marked equal, so the angle opposite the non-ticked side is 29°, and the angle $(11x-65)^\circ$ is the exterior angle? No, no, the angle $(11x-65)^\circ$ is an interior angle, marked with a curve, so it's the angle adjacent to the 29° angle. Wait, no: the triangle has a 29° angle, and two sides with ticks (equal sides), so the 29° angle is one base angle, and the angle $(11x-65)^\circ$ is the other base angle? No, no, the curve means it's the angle opposite the 29° angle? No, wait, the tick marks are on two sides: one side is between 29° and $(11x-65)^\circ$, and the other ticked side is opposite 29°. So the side opposite 29° is equal to the side between 29° and $(11x-65)^\circ$, so the angle opposite that side is $(11x-65)^\circ$, which equals 29°? No, no, exterior angle: the angle $(11x-65)^\circ$ is an exterior angle, so it equals the sum of the two remote interior angles. The two remote interior angles are 29° and the angle opposite the ticked side (which is equal to 29°, since sides are equal). So:

Step1: Exterior angle = sum of remote interiors

$11x - 65 = 29 + 29$

Step2: Solve for x

$11x = 58 + 65 = 123$
$x = \frac{123}{11} \approx 11.18$

Problem 11

Step1: $\triangle DBE$ is equilateral, so all angles = 60°, so $\angle 3 = \angle 4 = 60^\circ$, $\angle 5 = 60^\circ$. $\triangle ABC$ is isosceles, so $\angle A = \angle C$, so $4x+3 = 9x-47$

Step2: Solve for x

$3 + 47 = 9x - 4x$
$50 = 5x$
$x = 10$

Step3: Calculate $\angle A = \angle C = 4(10)+3 = 43^\circ$

Step4: $\angle 1 = \angle A = 43^\circ$, $\angle 9 = \angle C = 43^\circ$

Step5: $\angle 2 = 180 - \angle 1 - \angle 3 = 180 - 43 - 60 = 77^\circ$

Step6: $\angle 7 = 180 - \angle 9 - \angle 4 = 180 - 43 - 60 = 77^\circ$

Step7: $\angle 6 = 180 - \angle 7 - \angle 8$? No, wait, $\angle 6$ is in $\triangle BEC$, $\angle C=43^\circ$, $\angle 7=77^\circ$, so $\angle 6 = 180 - 43 - 77 = 60^\circ$? No, wait, no: $\angle 6$ is equal to $\angle 1$? No, wait, $\triangle ABC$ has $\angle ABC = 180 - 43 - 43 = 94^\circ$. $\angle ABC = \angle 2 + \angle 5 + \angle 7 = 77 + 60 + 77 = 214$? No, that's wrong. Wait, no: $\angle ABC = \angle 2 + \angle 5 + \angle 7$, and $\angle ABC = 180 - 2*43 = 94^\circ$. So $\angle 2 + \angle 5 + \angle 7 = 94$, and $\angle 5=60$, so $\angle 2 + \angle 7 = 34$. But $\angle 2 = \angle 7$ (since $\triangle ABD \cong \triangle CBE$), so $\angle 2 = \angle 7 = 17^\circ$. Oh right! I messed up: $\angle 3$ is $\angle ADB$, which is 60° (equilateral triangle), so in $\triangle ABD$, $\angle 1 + \angle 2 + \angle 3 = 180$, so $43 + \angle 2 + 60 = 180$, so $\angle 2 = 180 - 103 = 77$? No, no, $\triangle DBE$ is equilateral, so $BD=BE=DE$, and $\angle DBE=60^\circ$. $\triangle ABC$ is isosceles, so $AB=BC$. So $\angle ABD = \angle CBE = \frac{\angle ABC - 60}{2}$. $\angle ABC = 180 - 43 - 43 = 94$, so $\angle ABD = \angle CBE = \frac{94-60}{2}=17^\circ$. Ah, there we go! I misidentified $\angle 2$: $\angle 2$ is $\angle ABD$, not $\angle ADB$.

Correct Step3: $\angle ABC = 180 - 2*(4(10)+3) = 180 - 86 = 94^\circ$

Correct Step4: $\angle 5 = 60^\circ$ (equilateral triangle), so $\angle 2 = \angle 7 = \frac{94-60}{2}=17^\circ$

Correct Step5: In $\triangle ABD$, $\angle 1 = 43^\circ$, $\angle 2=17^\circ$, so $\angle 3 = 180 - 43 -17=120^\circ$

Correct Step6: $\angle 4 = \angle 3 = 120^\circ$ (equilateral triangle's exterior? No, $\triangle DBE$ is equilateral, so $\angle BDE=60^\circ$, so $\angle 3 = 180 - 60=120^\circ$, yes.

Correct Step7: $\angle 6 = \angle 1 = 43^\circ$, $\angle 8 = \angle 3 = 120^\circ$, $\angle 9 = \angle C=43^\circ$

Answer:

1. $x=4$
2. $x=\frac{205}{29} \approx 7.07$
3. $x=16$
4. $x=20.5$
5. $x=27$
6. $x=\frac{123}{11} \approx 11.18$

11.
$m\angle 1=43^\circ$, $m\angle 2=17^\circ$, $m\angle 3=120^\circ$, $m\angle 4=120^\circ$, $m\angle 5=60^\circ$
$m\angle 6=43^\circ$, $m\angle 7=17^\circ$, $m\angle 8=120^\circ$, $m\angle 9=43^\circ$