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# Measure of Angle | PRMO-2018 | Problem No-29

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

## Measure of Angle - PRMO 2018- Problem 29

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

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-29

#### Check the answer here, but try the problem first

$30$

## Try with Hints

#### First Hint

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?

#### Second Hint

$\angle E A D+\angle F A D=\angle E A F=\frac{A}{2}$
$\angle A E D=60^{\circ}+\frac{B}{2}$
$\angle A F D=\theta+\frac{C}{2}$
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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Try this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018.

## Measure of Angle - PRMO 2018- Problem 29

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

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-29

#### Check the answer here, but try the problem first

$30$

## Try with Hints

#### First Hint

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?

#### Second Hint

$\angle E A D+\angle F A D=\angle E A F=\frac{A}{2}$
$\angle A E D=60^{\circ}+\frac{B}{2}$
$\angle A F D=\theta+\frac{C}{2}$
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 $\quad \theta=30^{\circ}$

## Subscribe to Cheenta at Youtube

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