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Trigonometrical Ratios — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSTrigonometrical Ratios5 Marks
Question
In a right$-$angled $\triangle ABC, \angle B = 90^\circ , BD = 3, DC = 4$, and $AC = 13$. A point $D$ is inside the triangle such as $\angle BDC = 90^\circ .$
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Find the values of $3 - 2 \cos \angle BAC + 3 \cot \angle BCD$
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Answer

$\triangle B D C$ is a right$-$angled triangle.
$ \therefore B C^2$
$=B D^2+D C^2$
$=3^2+4^2$
$=9+16$
$=25$
$\Rightarrow B C=5 \ cm $
$\triangle A B C$ is a right$-$angled triangle.
$ \therefore A B^2$
$=A C^2-B C^2$
$=13^2-5^2$
$=169-25$
$=144$
$\Rightarrow A B=12 \ cm$
$3-2 \cos \angle B A C+3 \cot \angle B C D$
$=3-2 \times \frac{A B}{A C}+3 \times \frac{D C}{B D}$
$=3-2 \times \frac{4}{13}+3 \times \frac{4}{3}$
$=3-\frac{24}{13}+4$
$=7-\frac{24}{13}$
$=\frac{91-24}{13}$
$=\frac{67}{13} . $

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