Question

In triangle ABC, right angled at B,AB=15cm,BC=8cm. determine;. 1)sinA,cosA. 2)sinC,cosC

Answer 1

Given:

In triangle ABC, right-angled at B,

AB = 15 cm

BC = 8 cm

To determine:

(1) sinA, cosA

(2) sinC, cosC

Solution:

We have to first find the length of the hypotenuse i.e., AC by using the Pythagoras theorem:

$AC^2 = AB^2 + BC^2$

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

substituting the AB = 15 cm & BC = 8 cm

$\implies AC = \sqrt{15^2 + 8^2}$

$\implies AC = \sqrt{289}$

$\implies AC = 17\:cm$

(1). Finding the value of sin A and cos A:

From the figure attached below,

Considering the angle A, we have

BC = perpendicular = 8 cm

AC = hypotenuse = 17 cm

AB = base = 15 cm

∴ $\bold{sin \:A = \frac{perpendicular}{hypotenuse} = \frac{8}{17}}$

and

∴ $\bold{cos\:A = \frac{base}{hypotenuse} = \frac{15}{17}}$

(2). Finding the value of sin C and cos C:

From the figure attached below,

Considering the angle C, we have

BC = base = 8 cm

AC = hypotenuse = 17 cm

AB = perpendicular = 15 cm

∴ $\bold{sin \:C = \frac{perpendicular}{hypotenuse} = \frac{15}{17}}$

and

∴ $\bold{cos\:C = \frac{base}{hypotenuse} = \frac{8}{17}}$

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