Question

50. If in the triangle ABC the lengths of sides AB and BC are in the ratio of 13:5, find then value of cosec alpha
בם​

Answer 1

Step-by-step explanation:

AB / BC = 13 : 5

LET AB = 13X , BC= 5X

SO, BY USING PYTHAGORAS THEOREM

( AB )^2 = (AC) ^2 + (BC) ^2

(13X) ^2 =( AC )^2 + (5X)^2

(AC) ^ 2 = 169X^2 - 25 X ^2

AC = 12X

COSEC α = AC/BC

COSEC α = 12X/25X

COSEC α = 12

25

Answer 2

Given,

Ratio of length of sides of triangle,

$\[AB:BC = 13:5\]$

Solution,

Consider the triangle is ABC.

Use Pythagoras theorem for figure 1.

$\[\begin{array}{l}A{B^2} = A{C^2} + B{C^2}\\ \Rightarrow {\left( {13x} \right)^2} = A{C^2} + {\left( {5x} \right)^2}\\ \Rightarrow A{C^2} = 169{x^2} - 25{x^2}\\ \Rightarrow A{C^2} = 144{x^2}\\ \Rightarrow AC = 12x\end{array}\]$

Consider the angle A is $\[\alpha \]$.

$\[\cos ec\alpha = \frac{1}{{\sin \alpha }}\]$

$\[ \Rightarrow \cos ec\alpha = \frac{1}{{\frac{{BC}}{{AB}}}}\]$

$\[ \Rightarrow \cos ec\alpha = \frac{{AB}}{{BC}}\]$

$\[ \Rightarrow \cos ec\alpha = \frac{{13x}}{{5x}}\]$

$\[ \Rightarrow \cos ec\alpha = \frac{{13}}{5}\]$

Hence the $\[\cos ec\alpha \]$ is $\[\frac{{13}}{5}\]$.