Question

In triangles ABC, right angled at B, AC + BC = 25 cm and AB = 15 cm.
The value of sinA is:

Answer 1

AnsWer :

8/17.

Given :

• ∆ABC is a right angle triangle at B.
• Sum of, AC + BC = 25 cm.
• AB = 15 cm.

Solution :

• Let Side AC be x.

We equation become,

$\tt \implies AC + B = 25 \: cm.$

$\tt \implies x + BC = 25.$

$\tt \implies BC = 25 - x$

Now we have all sides of triangle,

• AB = 15 cm.
• AC = x cm.
• BC = 25 - x cm.

Figure Provide in attachment above.

Let find the value of x by apply Pythagoras theorem.

$\implies \tt {H}^{2} = {P}^{2} + {B}^{2} .$

$\implies \tt {x}^{2} = {(15)}^{2} + {(25 - x)}^{2} .$

$\implies \tt \cancel{x}^{2} = 225 + 625 + \cancel{x}^{2} - 50x.$

$\implies \tt 5 \cancel{0}x = 85 \cancel{0}.$

$\implies \tt x = \frac{85}{5}$

$\implies \tt x = 17 \: cm.$

Constant value of ∆ABC,

• AB = 15 Cm.
• BC = 8 Cm.
• AC = 17 Cm.

Now,

$\bigstar \: \tt\sin \theta = \frac{Perpendicular}{Hypotenuse}$

$\tt \implies \sin A= \frac{15}{17} .$

• Here angle take from Angle BAC or A.

Therefore , the value of Sin A is 8/17.