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AMC 10 Geometry Math Olympiad USA Math Olympiad

Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

Try this beautiful Problem on Geometry based on Measure of angle from AMC 10 A, 2014. You may use sequential hints to solve the problem.

Measure of angle  – AMC-10A, 2019- Problem 13


Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$

,

  • $90$
  • $100$
  • $105$
  • $110$
  • $120$

Key Concepts


Geometry

Circle

Triangle

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2019 Problem-`13

Check the answer here, but try the problem first

$110^{\circ}$

Try with Hints


First Hint

According to the questation we draw the diagram. we have to find out \(\angle BFC\)

Now \(\angle BEC\) = \(\angle BDC\) =\(90^{\circ}\) (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that \(\angle ABC=70^{\circ}\) (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

Second Hint

As \(\angle ABC=70^{\circ}\) and \(\angle BEC=90^{\circ}\) Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is\( 180^{\circ}\)

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

Third Hint

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$

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