Question

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm

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Answer 1

Answer:

If you want AC then,

AC^2 = AB^2 + AC^2.......…...by Pythagorean theorem

AC^2 = 24^2 + 7^2

= 576 + 49

= 625

AC = 25 ...... taking square root

therefore AC is 25cm

Answer 2

Corrected Question 1:

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

 

Step-by-step explanation:

 

Given:

A ΔABC where,

∠ABC = 90°

AB = 24cm

BC = 7cm

 

To find:

The values of sinA and cosA.

 

Solution:

We know that;

 

$\dashrightarrow \sf \ sinA = \dfrac{Side \ opposite \ to \ A}{Hypotenuse}$

 

$\dashrightarrow \sf \ cosA = \dfrac{Side \ adjacent \ to \ A}{Hypotenuse}$

 

In the given triangle ABC:

• Side opposite to ∠A is BC.
• Side adjacent to ∠A is AB.
• Hypotenuse is AC.

 

In ΔABC,

∠B = 90°

Using Pythagoras' Theorem;

 

➝ Hypotenuse² = Base² + Altitude²

➝ AC² = BC² + AB²

➝ AC² = (7)² + (24)²

➝ AC² = 49 + 576

➝ AC² = 625

➝ AC = √(625)

➝ AC = 25 cm.

 

Now, value of sinA is;

 

$\dashrightarrow \sf \ sinA = \dfrac{Side \ opposite \ to \ A}{Hypotenuse}$

 

$\dashrightarrow \sf \ sinA = \dfrac{BC}{AC}$

 

$\dashrightarrow \sf \ sinA = \bold{\dfrac{7}{25}}$

 

Now, value of cosA is;

 

$\dashrightarrow \sf \ cosA = \dfrac{Side \ adjacent \ to \ A}{Hypotenuse}$

 

$\dashrightarrow \sf \ cosA = \dfrac{AB}{AC}$

 

$\dashrightarrow \sf \ cosA = \bold{\dfrac{24}{25}}$

Hence solved.