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Ivenika [448]
3 years ago
13

Hurry up or ima get it wrong

Mathematics
2 answers:
ella [17]3 years ago
8 0

Angle 3 and angle 1 are vertical angles. This means that they are equal to each other. Since angle 3 is 111 degrees that means that angle 1 is also 111 degrees

Angle 2 is adjacent to angle 3. This means that their sum is 180 degrees. With this knowledge you can make a formula like so (let angle 2 be x):

111 + x = 180

x = 69

Angle 2 is 69 degrees

Angle 4 and angle 2 are adjacent. Again, this means that they are equal to each other. Angle 4 is 69 degrees

Hope this helped!

~Just a girl in love with Shawn Mendes

Nataly [62]3 years ago
4 0

Angle 1 and 3 are vertical angles, which mean they are identical. Since angle 3 is given as 111 degrees, then angle 1 is also 111 degrees.

All four angles need to equal 360.

Subtract angle 1 and 3 from 360:

360 - 111 - 111 = 138

Angle 2 and 4 would also be identical, so divide the 138 by 2:

138 / 2 = 69

Angle 2 and angle 4 are both 69 degrees.

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2 years ago
In triangle ΔABC, ∠C is a right angle and CD is the height to AB . Find the angles in ΔCBD and ΔCAD if: Chapter Reference a m∠A
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Answer:

Given A triangle ABC in which

 ∠C =90°,∠A=20° and CD ⊥ AB.

In Δ ABC

⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]

⇒20° + ∠B + 90°=180°

⇒∠B+110° =180°

∠B =180° -110°

∠B = 70°

In Δ B DC

∠BDC =90°,∠B =70°,∠BC D=?

∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]

90° + 70°+∠BC D =180°

∠BC D=180°- 160°

∠BC D = 20°

In Δ AC D

∠A=20°, ∠ADC=90°,∠AC D=?

∠A +  ∠ADC +∠AC D=180° [angle sum property of triangle]

20°+90°+∠AC D=180°

110° +∠AC D=180°

∠AC D=180°-110°

∠AC D=70°

So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°





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Answer:

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Step-by-step explanation:

Given that two points A and B on a plan represent two localities 12 m apart, to determine, given that the segment AB is 6 cm long, the scale used to draw the plan, the following calculation must be performed:

12 m = 12 cm x 100 = 1200 cm

1200/6 = 200

Therefore, the scale used to draw the plan is 1: 200.

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Lelechka [254]
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