Question

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

Answer 1

BC+CA=23
BC-CA  = 7
Adding these two equations
2BC=30
BC=15
Substituting this value in the first equation
15+CA=23
CA=8
$BC^{2}$+$AC^{2}$=$AB^{2}$ (by pythagorus theorem)
⇒(15*15)+(8*8)=$AB^{2}$
225+64=$AB^{2}$
$AB^{2}$=289
⇒AC=17

Answer 2

Given,

BC + CA = 23 cm

BC - CA = 7 cm

To find,

The value of Sin A and tan B.

Solution,

The value of Sin A is 15/17 and that of tan B is 8/15.

We can easily solve this problem by following the given process.

Now,

We know that the value of Sin and tan can only be found if we know the values of the base, perpendicular and hypotenuse.

According to the question,

BC + CA = 23 cm ---- equation 1

BC - CA = 7 cm ----- equation 2

Adding these two equations, we will get

2BC = 30 cm

BC = 30/2 cm

BC = 15 cm

Putting the value of BC in equation 1,

BC + CA = 23 cm ---- equation 1

15 + CA = 23 cm

CA = (23 - 15) cm

CA = 8 cm

Using the Pythagoras theorem in ∆ABC,

AB² = BC² + CA²

AB² = (15)² + (8)²

AB² = 225 + 64

AB² = 289

AB = √289

AB = 17 cm

So, for angle A,

The perpendicular will be BC and the hypotenuse will be AB.

We know that sin is perpendicular/hypotenuse.

Therefore,

Sin A = BC/AB

Sin A = 15/17

For angle B,

The perpendicular will be CA and the base will be BC.

Therefore,

Tan B = CA/BC

Tan B = 8/15

Hence, Sin A is 15/17 and Tan B is 8/15.