Hurry with math circles and tangents

m∠DAB = 90°

Step-by-step explanation:

* Lets revise some facts about the circle

- If two tangent segments drawn from a point outside the circle, then

 the two tangent segments are equal in length

- The radius and the tangent perpendicular to each other at the point

 of contact

* Lets solve the problem

∵ AB and AD are two tangent segments to circle C at B and D

 respectively

∴ AB = AD ⇒ (1)

∵ CD and CD are two radii

∴ AB ⊥ BC and AD ⊥ DC

∴ m∠ABC = m∠ADC = 90°

∵ m∠BDC = 45°

∵ ∠BDC + m∠ADB = m∠ADC

∴ 45° + m∠ADB = 90° ⇒ subtract 45° from both sides

∴ m∠ADB = 45° ⇒ (2)

- In Δ ABD

∵ AB = AD ⇒ proved in (1)

∴ m∠ABD = m∠ADB ⇒ isosceles triangle

∵ m∠ADB = 45° ⇒ proved in (2)

∴ m∠ABD = 45°

- The sum of the measure of the interior angles of a Δ is 180°

∴ m∠DAB + m∠ABD + m∠ADB = 180°

∴ m∠DAB + 45° + 45° = 180° ⇒ simplify

∴ m∠DAB + 90° = 180° ⇒ subtract 90° from both sides

∴ m∠DAB = 90°

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right