<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Answer:
tan2θ = 4√2/7
Step-by-step explanation:
Given sin theta=1/3 and 0 < theta< π/+
Required
tan 2 theta
tan2 theta = 2tanθ/1-tan²θ
Get tan θ
sinθ = opp/hyp
adj = √3²-1²²
adj = √9-1
adj = √8
tanθ = opp/adj = 1/2√2
tan2 theta = 2(1/2√2/1-(1/2√2)²
tan2θ = 1/√2/1-1/8
tan2θ = 1/√2/7/8
tan2θ = 8/7√2
Rationalize
tan2θ = 8√2/14
tan2θ = 4√2/7
Answer:
x = ±4
Step-by-step explanation:
Step 1: Write out quadratic
2x² - 5 = 27
Step 2: Add 5 to both sides
2x² = 32
Step 3: Divide both sides by 2
x² = 16
Step 4: Take the square root of both sides
x = ±4
∴ x can equal -4 or 4
Perpendicular lines, when intersecting, will always create a right angle at the intersection point. A right angle is an angle that is exactly 90 degrees.
