The population is tripling each minute
<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:
none
Step-by-step explanation:
We are given a triangle ABC with ∠A = 30°, sides a = 4 and b = 10.
According to the 'Law of Sines- Ambiguous Case', we have,
If a < b×sinA, then no triangle is possible.
If a = b×sinA, only one triangle is possible
If a > b×sinA, two triangles are possible.
So, we have,
b×sinA = 10 × sin30 = 10 × 0.5 = 5.
Now, as
4 = a < bsinA = 5.
We get, according to the rule, no triangle is possible.
I thinks it’s A or C )) but it’s not b or d, I’m pretty sure. So imma go with A or c
