In the given figure, BC = 15 cm and sin B = 4/5, show that

Given: BC =15cm and

We know that,

Or

Let,

Side opposite to angle B = 4k

and Hypotenuse = 5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AC)² + (BC)² = (AB)²

⇒ (4k)² + (BC)² = (5)²

⇒ 16k² + (BC)² = 25k²

⇒ (BC)² = 25 k² –16 k²

⇒ (BC)² = 9 k²

⇒ BC =√9 k²

⇒ BC =±3k

But side BC can’t be negative. So, BC = 3k

Now, we have to find the value of cos B and tan B

We know that,

Side adjacent to angle B = BC =3k

Hypotenuse = AB =5k

So,

Now, tan B

We know that,

side opposite to angle B = AC =4k

Side adjacent to angle B = BC =3k

So,

Now,

= –1 = RHS

Hence Proved