Question

*Consider a right triangle ABC, right angled at B. If AC = 10 units and BC = 8 units, determine Sin C.* 1️⃣ 4/3 2️⃣ 4/5 3️⃣ 3/4 4️⃣ 3/5​

Answer 1

Given:

Consider a right triangle ABC, right-angled at B. If AC = 10 units and BC = 8 units, determine Sin C.

To find:

Sin C

Solution:

In right-angled triangle ABC, we have

AC = 10 units

BC = 8 units

By using the Pythagoras theorem, we get

$\bold{AC^2 = AB^2 + BC^2}$

$\implies AB = \sqrt{AC^2 - BC^2}$

On substituting the values of AC = 10 and BC = 8, we get

$\implies AB = \sqrt{10^2 - 8^2}$

$\implies AB = \sqrt{100 - 64}$

$\implies AB = \sqrt{36}$

$\implies \bold{AB = 6\:units}$

We know that,

$\boxed{\bold{sin \:\theta = \frac{Perpendicular}{Hypotenuse} }}$

Therefore, by using the above trigonometric ratio for the right-angled triangle ABC, we get

$sin \:C = \frac{AB}{AC}$

on substituting AB = 6 units and AC = 10 units

$\implies sin \:C = \frac{6}{10}$

$\implies \bold{sin \:C = \frac{3}{5}}$ ← option (4)

Thus, sin C is → $\underline{\bold{\frac{3}{5}}}$ .

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