Question

$\Delta ABC$ is right angled at B, given $5\times sinA=3$. Find cos($\angle$C) + tan($\angle$A) + cosec($\angle$C).

• A. $\frac{12}{5}$
• B. $\frac{13}{12}$
• C. $\frac{13}{5}$
• D. $\frac{12}{5}$

Answer 1

The correct option is C $\frac{13}{5}$


$\Delta$ ABC is right angled at B,
Given,$5\times sinA=3sinA=\frac{3}{5}=\frac{BC}{AC}$
Which means BC=3k, AC=5k since BC:AC=3:5
Applying Pythagoras theorem, $A{B}^2+B{C}^2=A{C}^{2}A{B}^2=(5k{)}^2-(3k{)}^2=16k^{2}AB=4kcos\left(\angle C\right)=\frac{BC}{AC}=\frac{3}{5}tan\left(\angle A\right)=\frac{BC}{AB}=\frac{3}{4}cosec\left(\angle C\right)=\frac{AC}{AB}=\frac{5}{4}cos\left(\angle C\right)+tan\left(\angle A\right)+cosec\left(\angle C\right)=\frac{3}{5}+\frac{3}{4}+\frac{5}{4}=\frac{12+15+25}{20}=\frac{13}{5}$