Question

In $△ABC,m\angle A+m\angle C=m\angle B$ and $AC:AB=25:24$. if $BC=21$, find the area of $△ABC$.

Answer 1

$\measuredangle A+\measuredangle C=\measuredangle B$

also, $\measuredangle A+\measuredangle B+\measuredangle C={180}^0$

$\Rightarrow \measuredangle B={90}^0$

also, $x^2={21}^2+4^2$

$\Rightarrow \frac{625}{576}{y}^2={21}^2+y^2$

$\Rightarrow \frac{49}{576}{y}^2={21}^2$

$\Rightarrow \frac{7}{24}y=21\Rightarrow y=72$

$\Rightarrow x=75$

area $=\frac{1}{2}\times y\times 21=\frac{1}{2}\times 72\times 21=756$