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Explore the Back-StoryTry this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018.

Let $D$ be an interior point of the side $B C$ of a triangle ABC. Let $l_{1}$ and $l_{2}$ be the incentres of triangles $A B D$ and $A C D$ respectively. Let $A l_{1}$ and $A l_{2}$ meet $B C$ in $E$ and $F$ respectively. If $\angle B l_{1} E=60^{\circ},$ what is the measure of $\angle C l_{2} F$ in degrees?

,

- \(25\)
- \(20\)
- \(35\)
- \(30\)
- \(45\)

Trigonometry

Triangle

Angle

Pre College Mathematics

Prmo-2018, Problem-29

\(30\)

According to the questations at first we draw the picture . We have to find out the value of

$\angle C l_{2} F$. Now at first find out \(\angle AED\) and \(\angle AFD\) which are the exterioe angles of \(\triangle BEL_1\) and \(\triangle CL_2F\). Now sum of the angles is $180^{\circ}$

Now can you finish the problem?

$\angle E A D+\angle F A D=\angle E A F=\frac{A}{2}$

$\angle A E D=60^{\circ}+\frac{B}{2}$

$\angle A F D=\theta+\frac{C}{2}$

Therefore $\quad \ln \Delta A E F: \frac{A}{2}+60^{\circ}+\frac{B}{2}+\theta+\frac{C}{2}=180^{\circ}$

$90^{\circ}+60^{\circ}+\theta=180^{\circ}$ (as sum of the angles of a Triangle is $180^{\circ}$

Therefore $\quad \theta=30^{\circ}$

Try this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018.

Let $D$ be an interior point of the side $B C$ of a triangle ABC. Let $l_{1}$ and $l_{2}$ be the incentres of triangles $A B D$ and $A C D$ respectively. Let $A l_{1}$ and $A l_{2}$ meet $B C$ in $E$ and $F$ respectively. If $\angle B l_{1} E=60^{\circ},$ what is the measure of $\angle C l_{2} F$ in degrees?

,

- \(25\)
- \(20\)
- \(35\)
- \(30\)
- \(45\)

Trigonometry

Triangle

Angle

Pre College Mathematics

Prmo-2018, Problem-29

\(30\)

According to the questations at first we draw the picture . We have to find out the value of

$\angle C l_{2} F$. Now at first find out \(\angle AED\) and \(\angle AFD\) which are the exterioe angles of \(\triangle BEL_1\) and \(\triangle CL_2F\). Now sum of the angles is $180^{\circ}$

Now can you finish the problem?

$\angle E A D+\angle F A D=\angle E A F=\frac{A}{2}$

$\angle A E D=60^{\circ}+\frac{B}{2}$

$\angle A F D=\theta+\frac{C}{2}$

Therefore $\quad \ln \Delta A E F: \frac{A}{2}+60^{\circ}+\frac{B}{2}+\theta+\frac{C}{2}=180^{\circ}$

$90^{\circ}+60^{\circ}+\theta=180^{\circ}$ (as sum of the angles of a Triangle is $180^{\circ}$

Therefore $\quad \theta=30^{\circ}$

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