Question

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine the values of sinA and cosA.​

Answer 1

Step-by-step explanation:

${h}^{2} = {p}^{2} + {b}^{2} \\ {h}^{2} = {24}^{2} + {7}^{2} \\ {h}^{2} = 576 + 49 \\ {h}^{2} = 625 \\ h = \sqrt{625} \\ h = 25$

Answer 2

ANSWER :-

In Δ ABC, B is at right angle.

Given,

AB = 24cm

BC = 7cm

By using Pythagoras theorem,

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

⇒${(24)}^{2} + {(7)}^{2} = {AC}^{2}$

$⇒AC \: = \sqrt{ {(24)}^{2} + {(7)}^{2} } \\ = \sqrt{576 + 49} \\ = \sqrt{625}$

⇒AC = 25cm

Determine the values

$\sin \: A = \frac{BC }{AC} \\ \\ = \frac{7}{25}$

$\cos \: A = \frac{AB}{AC} \\ = \frac{24}{25}$

ANSWERED BY HELPINGSTUDENT.