Pre-Calculus - Solving Right Triangles

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

Given that:

C=90°

B=45°

a=5

c= $\sqrt{50}$

$a^{2}+b^{2}=c^{2}$

Therefore...

$5^{2}+b^{2}=(\sqrt{50})^{2}$

$25+b^{2}=50$

$b^{2}=50-25$

$b^{2}=25$

$\sqrt{b^{2}} =\sqrt{25}$

All angles of a triangle add up to 180°.

$A+45^{\circ}+90^{\circ}=180^{\circ}$

$A+135^{\circ}=180^{\circ}$

$A=180^{\circ}-135^{\circ}$

$A=45^{\circ}$

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

Given that:

C=90°

B=45°

a=5

c= $\sqrt{50}$

$a^{2}+b^{2}=c^{2}$

Therefore...

$5^{2}+b^{2}=(\sqrt{50})^{2}$

$25+b^{2}=50$

$b^{2}=50-25$

$b^{2}=25$

$\sqrt{b^{2}} =\sqrt{25}$

All angles of a triangle add up to 180°.

$A+45^{\circ}+90^{\circ}=180^{\circ}$

$A+135^{\circ}=180^{\circ}$

$A=180^{\circ}-135^{\circ}$

$A=45^{\circ}$

A right triangle has a base of 10 and a hypotenuse of 20. What is the length of the other leg?

$10\sqrt3$

$15\sqrt2$

$10\sqrt2$

$5\sqrt3$

Explanation

Write the Pythagorean Theorem.

Substitute the values of the leg and hypotenuse. The hypotenuse is the longest side of the right triangle. Solve for the unknown variable.

$a=\sqrt{300} = \sqrt{100\times3} = \sqrt{100}\times \sqrt3 = 10\sqrt3$

A right triangle has a base of 10 and a hypotenuse of 20. What is the length of the other leg?

$10\sqrt3$

$15\sqrt2$

$10\sqrt2$

$5\sqrt3$

Explanation

Write the Pythagorean Theorem.

Substitute the values of the leg and hypotenuse. The hypotenuse is the longest side of the right triangle. Solve for the unknown variable.

$a=\sqrt{300} = \sqrt{100\times3} = \sqrt{100}\times \sqrt3 = 10\sqrt3$

In the right triangle ABC, side AB is cm long, side AC is cm long, and side BC is the hypotenuse. How long is side BC?

$\sqrt{106}$ cm

$\sqrt{56}$ cm

Explanation

Given that ABC is a right triangle, the length of hypotenuse BC is the root of the sum of the squares of the two other sides (in other words, . Since AB is cm long and AC is cm long, we get that , and so $BC = \sqrt{106}$ .

In the right triangle ABC, side AB is cm long, side AC is cm long, and side BC is the hypotenuse. How long is side BC?

$\sqrt{106}$ cm

$\sqrt{56}$ cm

Explanation

Find the area of the given isosceles triangle:

Varsity log graph

$17.85; cm^{2}$

$49 ; cm^{2}$

$67.32 ; cm^{2}$

$35.7 ; cm^{2}$

$24.5; cm^{2}$

Explanation

The first step is to divide this isosceles triangle into 2 right triangles, making it easier to solve:

Varsity log graph

The equation for area is $A=\frac{1}{2}base\cdot height$

We already know the base, so we need to solve for height to get the area.

$tan20^{\circ}=\frac{height}{7; cm}$

$0.364=\frac{height}{7; cm}$

Then we plug in all values for the equation:

$Area=\frac{1}{2}\cdot14; cm\cdot2.55; cm=17.85; cm^{2}$

Find the area of the given isosceles triangle and round all values to the nearest tenth:

Varsity log graph

$51.1 ; cm^{2}$

$66.6 ; cm^{2}$

$133.2 ; cm^{2}$

$102.2 ; cm^{2}$

$46.2 ; cm^{2}$

Explanation

The first step to solve for area is to divide the isosceles into two right triangles:

Varsity log graph

From there, we can determine the height and base needed for our area equation $A=\frac{1}{2}\cdot b \cdot h$

$sin(67.5^{\circ})=\frac{1/2Base}{12; cm}$

$0.924=\frac{1/2Base}{12; cm}$

From there, height can be easily determined using the Pathegorean Theorem:

$(12; cm)^{2}=(11.1; cm)^{2}+h^{2}$

$h^{2}=144; cm^{2}-123.2; cm^{2}=20.8cm^{2}$

Now both values can be plugged into the Area formula:

$A=\frac{1}{2}\cdot22.2; cm\cdot4.6; cm=51.1cm^{2}$

In a right triangle, if the hypotenuse is and a leg is , what is the area of the triangle?

$\frac{3\sqrt7}{2}$

$3\sqrt{7}$

$\frac{15}{2}$

Explanation

Use the Pythagorean Theorem to find the other leg.

${}a^2+b^2=c^2$

$a=\sqrt7$

The length of the given leg is 3, and the unknown leg is $\sqrt7$ .

Use the area of a triangle formula and solve.

$A=\frac{bh}{2}= \frac{3\sqrt7}{2}$

Find the area of the given isosceles triangle:

Varsity log graph

$6.25 : cm^{2}$

$3.11 : cm^{2}$

$3.34 : cm^{2}$

$6.71 : cm^{2}$

$25.55: cm^{2}$

Explanation

The first step toward finding the area is to divide this isosceles triangle into two right triangles:

Varsity log graph

Trigonometric ratios can be used to find both the height and the base, which are needed to calculate area:

$sin(75^{\circ})=\frac{height}{5: cm}$

$0.9659=\frac{height}{5: cm}$

$sin(15^{\circ})=\frac{1/2base}{5: cm}$

$0.2588=\frac{1/2base}{5: cm}$

With both of those values calculated, we can now calculate the area of the triangle:

$Area=\frac{1}{2}\cdot2.59: cm\cdot4.83: cm=6.25: cm^2$

Solving Right Triangles

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