<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>
Congruent to angle 1: 3, 5, 10, 8, and 6
congruent to angle 2: 4, 9, and 7
Answer:
Step-by-step explanation:
5.24×10⁸ miles × (1 hour)/(3.2×10⁴ miles) = 1.6375×10⁴ hours = 16,375 hours
Answer:
y = 1/2x - 6
Step-by-step explanation:
y2 - y1 / x2 - x1
-2 - (-5) / 8 - 2
3 / 6
= 1/2
y = 1/2x + b
-5 = 1/2(2) + b
-5 = 1 + b
-6 = b
Answer: a) It will be a right triangle with a hypotenuse of 75 and a vertical leg that is 62.
b) The length of the kite string is 75 (given in the problem). However, I believe you are looking for the distance from the spot on the ground beneath the kite to Janet. That is about 42.2 m.
To find the missing distance in the right triangle. You have to use the Pythagorean Theorem. I set it up and solve it below.
a^2 + b^2 = c^2
x^2 + 62^2 = 75^2
x^2 + 3844 = 5625
x^2 = 1781
x = 42.2 (about)
