Question

in right angle triangle ABC,8cm,15cm and 17cm are the lengths of AB,BC and CA respectively. Then,find out sinA, cos A and tan A.​

Answer 1

$\large\underline{\sf {\underline{Given \: that}} : } ☞$

• In a right angle triangle ABC 8cm, 15cm, 17cm
• They are lengths of AB, BC, and, CA.

$\large\underline{\sf {\underline{To\;Find }} : } ☞$

• Find out sin A, cos A and tan A

$\large\underline{\sf {\underline{Need \;formulas }} : } ☞$

$\sf \: ➩Sin A = \frac {opposite \;side} { hypotenuse} \\ \\ ➩\sf \: Cos A = \frac{adjacent \; side }{ Hypotenuse } \\ \\ ➩\sf \: tan A = \frac {opposite\; side }{adjacent\; side }$

$\large\underline{\sf {\underline{solution }} : } ☞$

In a question it is clear that the in a triangle ABC which

AB = 8cm, BC = 15cm, AC = 17cm

According to the question

• AB = Adjacent side

• BC = Opposite side

• AC = Hypotenuse

Now we have to find sin A, cos A and tan A

Let find sin A first

We know that formula

$\sf Sin \;A = \frac {lengths\; of the \;side\; opposite \;to\; angle \;A }{ lengths \;of\; hypotenuse } \\ \\ \sf sin \: A = \frac{BC}{AC} \\ \\ \sf \:sin \:A = \frac{15}{17}$

________________________

Now we find cos A

$\sf \: Cos\;A = \frac{ lengths \;of \;opposite\; side \;of\; angle \;A}{ length\; of \;hypotenuse }</p><p> \\ \\ \sf \: cos\;A = \frac{AB}{AC}\\\\\sf\: cos \;A= \frac{8}{17}$

________________________

Now tan A

$\sf \: tan A =\frac{ length\; of\; adjacent \;side \;of \;angle\; A} {length \;of\; opposite\; side \;of\; \;angle\; A} \\ \\ \sf \: tan\;A = \frac{BC}{AB} \\ \\ \sf \: tan\;A = \frac{15}{8}$

_______________________

Hence,

The value of sin A = 15/17, cos A = 8/17 and tan A = 15/8

____________

--------------------------------

More information

Theorem: In a right-angled triangle, if the sum of the square of two sides is equal to the square of one side, then the right angle is the angle that is opposite to the first side.