Question

In triangle ABC, $AB=5,BC=12,AC=13,\angle ACB=x$ and $\angle ABC={90}^{\circ }.$ If the value of $\sin x+\sec x+\tan x$ is $a,$ then $156a=$

• A. 394
• B. 384
• C. 294
• D. 400

Answer 1

The correct option is C 294

The triangle is:

For angle $x$ in the right-angled triangle,

$\sin x=\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\sec x=\frac{\text{Hypotenuse}}{\text{Opposite side}}=\frac{1}{\cos x}$

$\tan x=\frac{\text{Opposite side}}{\text{Adjacent side}}$

From this triangle, we get

$\sin x=\frac{5}{13}$

$\sec x=\frac{13}{12}$

$\tan x=\frac{5}{12}$

$\Rightarrow \sin x+\sec x+\tan x=\frac{5}{13}+\frac{13}{12}+\frac{5}{12}$

$\Rightarrow a=\frac{294}{156}$

$\Rightarrow 156a=294$