Question

In a right angle triangle ABC,right angle is at C. BC + CA= 23 cm and BC-CA=7 cm,
then find sin A and tan B.​

Answer 1

Step-by-step explanation:

BC+CA=23

BC=23-CA

BC-CA=7

23-CA-CA=7

-2CA=7-23

-2CA=-16

CA=8

BC=23-CA

BC=23-8

BC=15

AB^2=BC^2+CA^2

AB^2=225+64

AB^2=289

AB=17

sinA=p/h

=15/17

tanB=p/b

=17/15

Answer 2

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Given,

2x2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 2, b = -3 and c = 5

We know, Discriminant = b2 – 4ac

= ( – 3)2 – 4 (2) (5) = 9 – 40

= – 31

As you can see, b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0.