Question

The triangle ABC is an equilateral triangle. Given m∠B = 3y, and m∠C = 2x, find the value of x and y.

Answer 1

Answer:-

Given that ΔABC is an equilateral triangle

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘ ⇒∠x+46∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘ ⇒∠x+46∘ =60∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘ ⇒∠x+46∘ =60∘

Given that ΔABC is an equilateral triangle⇒∠A=∠B=∠C=60 ∘ In a ΔDAB,DA=DB ....Given⇒∠ABD=∠DAB angles opposite to equal sides are also equal Sum of all angles of a triangle =180 ∘ .⇒∠ABD+∠DAB+88∘ =180∘ ⇒∠ABD+∠DAB=92∘ ⇒∠ABD=∠DAB=46∘ ⇒∠B=∠x+∠ABD=60∘ ⇒∠x+46∘ =60∘ ⇒∠x=14∘

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