Question

In a ∆ABC, right angled at 'B' AB = 15 cm, BC = 8cm. Then sin A =

plzz help me with explanation ​

Answer 1

Using Pythagoras theorem in △ABC

(AB)

2

+(BC)

2

=(AC)

2

(15)

2

+(8)

2

=(AC)

2

⇒(AC)

2

=225+64=289

⇒AC=17cm

Now inradius r=

s

Δ

where Δ is the area of triangle and s is semi perimeter.

Δ=

2

1

×8×15=60

s=

2

8+15+17

=

2

40

=20

⇒r=

s

Δ

=

20

60

=3cm

Answer 2

Step-by-step explanation:

In ∆ ABC

angle B=90°

by Pythagoras theorem

${ac}^{2} = {ab}^{2} + {bc}^{2} \\ {ac }^{2} = {15}^{2} + {8}^{2} \\ {ac}^{2} = 225 + 64 \\ {ac}^{2} = 289 \\ taking \: square \: root \: on \: both \: sides \\ ac = 17 \\ therefore \: \\ sin \: a = \frac{opposite \:side \: }{hypotenuse} \\ = \frac{bc}{ac} \\ = \frac{8}{7} \\ therefore \: \sin(a) = \frac{8}{7}$

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