For the first one you would do:
X rad 3=7 divide by rad 3 in both sides and you don’t wanna leave a radical in the denominator so multiply rad 3 in both denom. And numerator and then you get 7rad3/3.
A 30:60:90 triangle always have n:2n:n rad 3 so since 2n=10 you would get n=5 and since BC is n rad3 so you would get 5rad 3.
In a 45:45:90 triangle it is n:n:n rad 2, since AC=3 then AB= 3 and then BC= 3 rad 2
Since AC and AB are congruent you they both would be 12 so it is true.
Answer:
It’s the second one
Step-by-step explanation:
Answer: The proof is mentioned below.
Step-by-step explanation:
Here, Δ ABC is isosceles triangle.
Therefore, AB = BC
Prove: Δ ABO ≅ Δ ACO
In Δ ABO and Δ ACO,
∠ BAO ≅ ∠ CAO ( AO bisects ∠ BAC )
∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )
BO ≅ OC ( O is the mid point of BC)
Thus, By ASA postulate of congruence,
Δ ABO ≅ Δ ACO
Therefore, By CPCTC,
∠B ≅ ∠ C
Where ∠ B and ∠ C are the base angles of Δ ABC.