Question

On a piece of paper, use a protractor to construct right triangle ABC with AB=3 in. , m∠A=90° , and m∠B=45° .

Answer 1

Answer:

AC = 3 in.

Step-by-step explanation:

It is given in the figure attached, in Δ ABC

∠ A = 90°

m∠B = 45°

AB = 3 in.

We have to find all sides of the triangle given

we will apply sine in Δ ABC to find the measure of AC

sinB = $\frac{AB}{BC}$

$sin45=\frac{3}{BC}$

$\frac{1}{\sqrt{2}}=\frac{3}{BC}$

Now by cross multiplication

BC = 3√2 in.

Now by Pythagoras theorem

BC²= AC² + AB²

(3√2)² = AC² + 3²

18 - 9 = AC²

AC = √9 = 3

Now we come to the options you have mentioned

AC = 3 in. is the correct answer.

Answer 2

Answer:

AC=3 in, m∠C=45° and BC=3√2 in.

Step-by-step explanation:

Given information: ABC is a right angled triangle AB=3 in. , m∠A=90° , and m∠B=45° .

According to the angle sum property the sum of interior angles of a triangle is 180°.

$\angle A+\angle B+\angle C=180^{\circ}$

$90^{\circ}+45^{\circ}+\angle C=180^{\circ}$

$135^{\circ}+\angle C=180^{\circ}$

$\angle C=180^{\circ}-135^{\circ}$

$\angle C=45^{\circ}$

The measure of angle C is 45°.

In a right angled triangle,

$\tan\theta=\frac{opposite}{adjacent}$

In triangle ABC,

$\tan B=\frac{AC}{AB}$

$\tan (45^{\circ})=\frac{AC}{3}$

$1=\frac{AC}{3}$

Multiply both sides by 3.

$3=AC$

The measure of AC is 3 in.

According to Pythagoras theorem,

$hypotenuse^2=base^2+perpendicular^2$

$BC^2=3^2+3^2$

$BC^2=9+9$

$BC^2=18$

Taking square root on both sides.

$BC=\sqrt{18}$

$BC=3\sqrt{2}$

Therefore, the missing measurements are AC=3 in, m∠C=45° and BC=3√2 in.