Question

In a right angled triangle ABC right angled at B,
$5\times sinA=3$.
Find the value of cos C + tan A + cosec C. [3 Marks]

Answer 1

Given, triangle ABC is right angled at B.
$\text{Also,}5\times sinA=3$

$\Rightarrow sinA=\frac{3}{5}=\frac{BC}{AC}$ [1 Mark]

Let BC = 3k, AC = 5k $(\because BC:AC=3:5)$

Applying Pythagoras theorem,
$A{B}^2+B{C}^2=A{C}^{2}A{B}^2=(5k{)}^2-(3k{)}^2=16k^{2}AB=4k$

$cosC=\frac{BC}{AC}=\frac{3}{5}tanA=\frac{BC}{AB}=\frac{3}{4}cosecC=\frac{AC}{AB}=\frac{5}{4}$
[1 Mark]

Hence,
$cosC+tanA+cosecC=\frac{3}{5}+\frac{3}{4}+\frac{5}{4}=\frac{12+15+25}{20}=\frac{13}{5}$

[1 Mark]