Question

In the right △ABC with m∠C=90°, m∠B=75°, and AB = 12 cm. Find the area of △ABC.

Answer 1

Given that ABC is a right angle triangle with m∠C=90°,  m∠B=75° and AB=12cm.

Area of triangle =  

                         =  

So we need to find BC,AC now.

To find them let us take sinB and cosB.

SinB =  

BC = 12* cos B = 12* cos 75

So, area =  

            = 72*sin(75)*cos(75)

            = 36*(2sin(75)cos(75))

            = 36*sin(150) =  18

Hence area of triangle is 18

Hope this helps tbh sorry I was a little late btw (:

Answer 2

The area of the given right-angle  is 18 cm².

What is a right-angle triangle?

A right-angle triangle is a triangle that has a right angle as an interior angle.

Given, right △ABC with m∠C=90°, m∠B=75°, and AB = 12 cm.

Therefore, sin (∠B) = $\frac{AC}{AB}$

⇒ sin(75°) = $\frac{AC}{12}$

⇒ AC = 12 × sin(75°)

⇒ AC = 12 × 0.966

⇒ AC = 11.59 cm

Again, tan(∠B) =  $\frac{AC}{BC}$

⇒ tan(75°) = $\frac{11.59}{BC}$

⇒ 3.732 = $\frac{11.59}{BC}$

⇒ BC = $\frac{11.59}{3.732}$ cm = 3.105 cm

Therefore, area of the ΔABC

= $\frac{1}{2}$ × AC × BC

=  $\frac{1}{2}$ × 11.59 × 3.105 cm²

= 18 cm²

Learn more about a right-angle triangle here: https://brainly.com/question/11108818

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