Point E is exterior to APD
Answer:
The answer is 2
Step-by-step explanation:
Multiply the numerator and denominator by the conjugate.
Given:
In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.
To find:
The measure of angle P.
Solution:
According to the Law of Cosines:
Using Law of Cosines in triangle OPQ, we get
On further simplification, we get
Therefore, the measure of angle P is 79 degrees.
I think the awkward science is quite amusing to me
<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>
