Question

in a right angled triangle ABC, right angled at A, BC= 41cm and AB-AC =31cm. Find all the trigonometry ratios

Answer 1

Consider the right angled triangle ABC, right angled at A.

By applying Pythagoras theorem, we get

$BC^2 = AC^2 + AB^2$   (Equation 1)

Since, AB-AC = 31

So, AB = AC + 31

Substituting the value of AB in equation 1, we get

$(41)^2 = (31+AC)^2 +(AC)^2$

$1681 = 961+AC^2+62AC+AC^2$

$1681 = 961+2AC^2+62AC$

$2AC^2+62AC-720 = 0$

$AC^2+31AC-360 = 0$

AC = $\frac{-31\pm\sqrt{31^2+4(360)}}{2}$

AC=$\frac{-31\pm\sqrt{2401}}{2}$

AC = $\frac{-31\pm49}{2}$

Since,the measure of segment AC can not be negative.

So,AC = 9 cm

AB-AC = 31

AB - 9 = 31

AB = 40 cm

Now, we will determine all the trigonometric ratios.

Let us find all the ratios about angle B.

$\sin B = \frac{P}{H} = \frac{9}{41}$  ,$\csc B= \frac{41}{9}$

$\cos B = \frac{B}{H} = \frac{40}{41}$, $\sec B= \frac{41}{40}$

$\tan B = \frac{P}{B} = \frac{9}{40}$, $\cot B= \frac{40}{9}$

Let us find all the ratios about angle C.

$\sin C = \frac{P}{H} = \frac{40}{41}$  ,$\csc C= \frac{41}{40}$

$\cos C = \frac{B}{H} = \frac{9}{41}$, $\sec C= \frac{41}{9}$

$\tan C = \frac{P}{B} = \frac{40}{9}$, $\cot C= \frac{9}{40}$