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 = $\frac{1}{2} * base * height$

                          = $\frac{1}{2} *BC*AC$

So we need to find BC,AC now.

To find them let us take sinB and cosB.

SinB = $\frac{AC}{AB} = \frac{AC}{12}$

$AC = 12*sinB = 12*sin(75)$

$cos B = \frac{BC}{AB} = \frac{BC}{12}$

BC = 12* cos B = 12* cos 75

So, area = $\frac{1}{2} * 12 * sin(75)*12* cos(75)$

             = 72*sin(75)*cos(75)

             = 36*(2sin(75)cos(75))

             = 36*sin(150) =  18

Hence area of triangle is 18 $cm^{2}$

Answer 2

The area of $\triangle\text{ ABC}$ is $\boxed{\bf 18\text{\bf\ square units}}$.

Further explanation:

Given information:

In $\triangle\text{ ABC}$ the measure of $\angle\text{C}$ is $90^{\circ}$and measure of $\angle\text{B}$ is $75^{\circ}$.

The length of AB is $12\text{ cm}$.

The measure of $\angle\text{A}$ can be calculated as follows:

$\begin{aligned}\angle\text{A}&=180^{\circ}-(\angle\text{B}+\angle\text{C})\\&=180^{\circ}-(75^{\circ}+90^{\circ})\\&=180^{\circ}-165^{\circ}\\&=15^{\circ}\end{aligned}$

The $\triangle\text{ ABC}$ is shown in Figure 1.

Consider the length of the perpendicular of $\triangle\text{ ABC}$ as $x$ and the length of the base as $y$.

The perpendicular length $x$ is calculated as follows:

$\begin{aligned}\sin 75^\circ&=\frac{x}{{12}}\\ \frac{{\sqrt3+1}}{{2\sqrt2}}&=\frac{x}{{12}}\\ \frac{{\sqrt3+1}}{{2\sqrt2}}\times12&=x\\ \frac{{6\left({\sqrt3+1}\right)}}{{\sqrt2}}\times\frac{{\sqrt2}}{{\sqrt2}}&=x\\ x&=3\sqrt 2\left({\sqrt3+1}\right)\\ \end{aligned}$

The base length $y$ is calculated as follows:

$\begin{aligned}\sin 15^\circ&=\frac{y}{{12}}\\\frac{{\sqrt3-1}}{{2\sqrt2}}&=\frac{y}{{12}}\\ \frac{{\sqrt3-1}}{{2\sqrt2}}\times12&=y\\ \frac{{6\left({\sqrt3-1}\right)}}{{\sqrt2}}\times\frac{{\sqrt 2}}{{\sqrt 2}}&=y\\ y&=3\sqrt2\left({\sqrt3-1}\right)\\ \end{aligned}$

Area of right angled triangle is calculated as follows:

$\boxed{\text{Area}=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Perpendicular}}$

The length of base of the triangle is $3\sqrt{2}(\sqrt{3}-1)$ units and length of perpendicular is $3\sqrt{2}(\sqrt{3}+1)$ units.

Now, the area of $\triangle\text{ ABC}$ is calculated as follows:

$\begin{aligned}\text{Area}&=\dfrac{1}{2}\cdot \left[3\sqrt{2}(\sqrt{3}-1)\right]\cdot \left[3\sqrt{2}(\sqrt{3}+1)\right]\\&=\dfrac{1}{2}\cdot \left[9\cdot 2\left((\sqrt{3})^{2}-1\right)\right]\\&=\dfrac{1}{2}\cdot 18\cdot 2\\&=18 \end{aligned}$

Thus, the area of $\triangle\text{ ABC}$ is $\boxed{\bf 18\text{\bf\ square units}}$.

Learn more:

1. Learn more about the division: brainly.com/question/545132

2. Learn more about the simplification brainly.com/question/894273

3. Learn more about the percentage brainly.com/question/928382

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Triangles

Keywords: Triangle, right-angled triangle, sin75, sin15, 12cm, ABC, area, area of triangle ABC, m \angle C=90, m \angle B=75, perpendicular, base.