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?
,
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?
,
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 $\quad \theta=30^{\circ}$
