Question

PLEASE HELP me find the rest of the angles. The big triangle problem

Answer 1

Answer:

∠1 = 100°

∠2 = 56°

∠3 = 128°

∠4 = 36°

∠5 = 94°

∠6 = 40°

∠7 = 30°

∠8 = 60°

∠9 = 26°

∠10 = 38°

∠11 = 16°

∠12 = 30°

∠13 = 60°

∠14 = 110°

∠15 = 90°

∠16 = 34°

∠17 = 86°

∠18 = 52°

∠19 = 120°

∠20 = 70°

Step-by-step explanation:

Before we solve for all the angles, first we figure out what we can with the given.

Just remember the following:

• The sum of all angles in any triangle is always 180°.

• The sum of complementary angles (two angles that form a right angle) is 90°.

• The sum of supplementary angles (two angles that form a straight line) is 180°

• An isosceles triangle which means that two sides are equal and as a result, two angles are equal.
• Two line segments that are perpendicular to each other create a right angle or an angle with a measure of 90°.
• Vertical angles are congruent.

GIVEN: ΔABC is an isosceles triangle

∠A = ∠1 = 100°

∠B = ∠6

∠B = ∠C

100° + 2(∠6) = 180°

∠6 = 80°/2

∠6 = 40°

GIVEN: ΔDEF is an equilateral triangle. We can then assume that

∠13 = 60°

∠8 = 60°

GIVEN: ∠13 is supplementary to ∠19 based on the figure

Then we can assume that:

∠13 + ∠19 = 180°

60° + ∠19 = 180°

∠19 = 120°

GIVEN: AE ⊥ EF then we can assume that the create a right angle. This means that the measure of the angle the create or m∠AEF = 90°.

Since ∠7 is complementary to the angle of the equilateral triangle ΔDEF

we can assume that

∠7 + 60° = 90 °

∠7 = 30°

GIVEN: ∠7, ∠19 and ∠12 make a triangle based on the figure:

We can then assume that:

∠7 +  ∠19 +∠12 = 180°

Based on what we solved above:

30° + 120 + ∠12 = 180°

150° + ∠12 = 180°

∠12 = 30°

GIVEN: CG ⊥ EG

We can then assume that m∠EGC = 90°

∠ECG = ∠15

∠15 = 90°

GIVEN: ∠3 = 128 and ∠4 = 36

Notice that ∠4 and the vertical angle of ∠3 make a triangle with ∠11. Because ∠3 is the vertical of the angle that completes the triangle then we know that it is equal to 128° as well.

We can then assume:

∠4 + vertical of ∠3 + ∠11 = 180°

36° + 128° + ∠11 = 180°

164° + ∠11 = 180°

∠11 = 16°

GIVEN: ∠5 = 94°

Notice that ∠5, ∠11 and ∠20 make a triangle. We already know that:

∠11 = 16° from the previous solution.

∠5 + ∠11 + ∠20 = 180°

94° + 16° + ∠20 = 180°

110° + ∠20 = 180°

∠20 = 70°

GIVEN: ∠12, ∠6 and ∠CBD , make a triangle, where:

∠6 = 60°

∠12 = 30°

∠12 + ∠6 + ∠CBD = 180°

60° + 30° + ∠CBD = 180°

90° + ∠CBD = 180°

∠CBD = 90°

Now look at the figure and you'll notice that:

∠CBD, ∠2 and ∠16 make a triangle

∠2 = 56° (given)

∠CBD = 90°

∠CBD + ∠2 + ∠16 = 180°

90 + 56 + ∠16 = 180

146 + ∠16 = 180

∠16 = 34°

GIVEN: ∠19, ∠16, and ∠CDG make a straight line

∠19 + ∠16 + ∠CDG = 180°

120° + 34° + ∠CDG = 180°

154° + ∠CDG = 180°

∠CDG = 26°

Now see that ∠CDG, ∠3 and ∠9 make a triangle:

∠CDG + ∠3 + ∠9= 180°

26 + 128 + ∠9 = 180

154 +∠9 = 180

∠9 = 26°

GIVEN: ∠20 is supplementary to ∠14 as shown on the figure.

We can assume then that:

∠20 + ∠14 = 180°

70° + ∠14 = 180°

∠14 = 110°

GIVEN: ∠3 is supplementary to ∠18, where ∠3 = 128°

∠3 + ∠18 = 180°

128 + ∠18 = 180°

∠18 = 52°

GIVEN: ∠18, ∠10 and ∠15 make a triangle

∠18 +∠10 + ∠15 = 180

52 + ∠10 + 90 = 180

142 + ∠10 = 180

∠10 = 38°

AND LASTLY: ∠4, ∠D and ∠14 make a triangle

∠4 + ∠D + ∠14 = 180°

36 + ∠D + 110 = 180

146 + ∠D = 180

∠D = 34°

Now in line GE, ∠13, ∠17 and ∠D make a straight line

∠13 + ∠17 + ∠D = 180°

60 + ∠17 + 34 = 180

94 + ∠17 = 180

∠17 = 86°

Answer 2

Refer the below solution for better understanding.

Step-by-step explanation:

Given :

Triangle ABC is an isosceles triangle.

Triangle DEF is an equilateral triangle.

$\angle$13 is supplementary to $\angle$19 based on the figure.

AE $\perp$ EF

CG $\perp$ EG

$\angle 3 = 128^\circ\;and \; \angle4 = 36^\circ$

$\angle 5 = 94^\circ$

$\angle$20 is supplementary to $\angle$14.

$\angle$3 is supplementary to $\angle$18.

Solution :

Given that triangle ABC is an isosceles triangle. So

$\rm \angle A = \angle 1 = 100^\circ$

$\rm \angle B = \angle 6$

$\rm \angle B=\angle C$

$100^\circ + 2(\angle 6)=180^\circ$

$\angle 6 = 40^\circ$

Given that triangle DEF is an equilateral triangle. So,

$\angle 13 = 60^\circ$

$\angle 8 = 60^\circ$

$\rm \angle 13+ \angle 19 = 180^\circ$      ($\angle$13 is supplementary to $\angle$19)

$60^\circ + \angle 19 = 180^\circ$

$\angle 19 = 120^\circ$

$\angle 7 + 60^\circ = 90^\circ$    (AE $\perp$ EF)

$\angle 7 = 30^\circ$

$\angle 7 +\angle 19+\angle 12 = 180^\circ$

$30^\circ + 120^\circ + \angle 12 = 180^\circ$

$\angle 12 = 30^\circ$

$\rm \angle ECG = \angle 15$      (CG $\perp$ EG)

$\angle 15 = 90^\circ$

$\angle 4 +\angle 3+\angle 11 = 180^\circ$

$36^\circ+128^\circ+\angle 11 = 180^\circ$      (Given : $\angle 3 = 128^\circ\;and \; \angle4 = 36^\circ$)

$\angle 11 = 16^\circ$

$\angle 5 +\angle 11+\angle 20 = 180^\circ$  

$94^\circ + 16^\circ + \angle 20 = 180^\circ$            (Given : $\angle 5 = 94^\circ$)

$\angle 20 = 70^\circ$

$\rm \angle 12 + \angle 6 + \angle CBD = 180^\circ$        (Triangle)

$\rm 60^\circ+30^\circ+\angle CBD= 180^\circ$

$\rm \angle CBD = 90^\circ$

$\rm \angle CBD + \angle 2 + \angle 16 = 180^\circ$        (Triangle)

$90^\circ + 56^\circ + \angle 16 = 180^\circ$

$\angle 16 = 34^\circ$

$\rm \angle 19 + \angle 16 +\angle CDG = 180^\circ$        (Straight Line)

$\rm \angle CDG = 26^\circ$

$\rm \angle CDG + \angle 3 + \angle 9 = 180^\circ$            (Triangle)

$26^\circ + 128^\circ +\angle 9 = 180^\circ$

$\angle 9 = 26^\circ$

$\angle 3 + \angle 18 = 180^\circ$               ($\angle$3 is supplementary to $\angle$18)

$128^\circ +\angle 18 = 180^\circ$

$\angle 18 = 52^\circ$

$\angle 18 + \angle10 +\angle15 = 180^\circ$        (Triangle)

$52^\circ + \angle 10 + 90^\circ = 180^\circ$

$\angle 10 = 38^\circ$

$\rm \angle 4 + \angle D + \angle 14 = 180^\circ$           (Triangle)

$\rm 36^\circ+ \angle D + 110^\circ = 180^\circ$

$\rm \angle D = 34^\circ$

$\rm \angle 13 +\angle 17 + \angle D = 180^\circ$          (Straight Line)

$60^\circ + \angle 17 + 60^\circ = 180^\circ$

$\angle 17 = 86^\circ$

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