Question

4. Given 15 cot A = 8, find sin A and sec A.​

Answer 1

15 cot A= 8

cot A=15/8

base/perpendicular=15/8

base=15

perpendicular=8

By Pythagoras theorem

(8)²+(15)²=(hypotenuse)²

(hypotenuse)²=64+225

(hypotenuse)²=289

(hypotenuse)=17

Now,

sinA=perpendicular/hypotenuse

=8/17

secA=hypotenuse/base

=17/15

Hope it helps

:-)

Answer 2

Hello Mate,

Here is your answer,

Given:

$15 \cot\alpha = 8 \\ \\ \cot \alpha = \frac{8}{15}$

We Know,

$cot \alpha = \frac{adjacent \: side}{opposite \: side} = \frac{8}{15}$

So we can say that:-

The adjacent side is 8 units.

The opposite side is 15 units.

We have to find the value of hypotenuse

${a}^{2} + {b}^{2} = {c}^{2} \\ \\ {8}^{2} + {15}^{2} = {c}^{2} \\ \\ {c}^{2} = 225 + 64 \\ \\ {c}^{2} = 289 \\ \\ c = \sqrt{289} \\ \\ c = hypotenuse = 17 \: units$

Now as we know the value of hypotenuse as 17 units we can now find out

$\sin\alpha \: \: \: and \: \: \: \sec \alpha$

We Know,

$\sin \alpha = \frac{opposite \: side}{hypotenuse} \\ \\ \sin \alpha = \frac{8}{17} \\ \\ and \\ \\ \sec \alpha = \frac{hypotenuse}{adjacent \: side} \\ \\ \sec \alpha = \frac{17}{15}$

Hope this solution helps:)

Plz follow:)