Question

What is sinA in right triangle ABC, where C is the right angle, AC=6, and AB=10

Answer 1

Answer:

$SinA=\frac{4}{5}$

Step-by-step explanation:

Given,

        In the Right angled triangle ABC.

            $\angle C=90^0\\ \\ AC=6\\ \\ AB=10$

According to Pythagoras' theorem the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

                            $h^2=b^2+p^2\\ \\ AB^2=AC^2+BC^2\\ \\ 10^2=6^2+AB^2\\ \\ AB^2=10^2-6^2\\ \\ AB^2=100-36\\ \\ AB^2=64\\ \\ AB=\sqrt{64}=8$

we know that,

                          $SinA=\frac{BC}{AB}\\ \\ SinA=\frac{p}{h} \\ \\ SinA=\frac{8}{10} \\ \\ SinA=\frac{4}{5}$