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Precalculus : Solve a Right Triangle

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Example Questions

Example Question #11 : Solve A Right Triangle

Find the area of the given isosceles triangle:

Varsity log graph

Possible Answers:

$24.5\; cm^{2}$ $\displaystyle 24.5\; cm^{2}$

$67.32 \; cm^{2}$ $\displaystyle 67.32 \; cm^{2}$

$49 \; cm^{2}$ $\displaystyle 49 \; cm^{2}$

$35.7 \; cm^{2}$ $\displaystyle 35.7 \; cm^{2}$

$17.85\; cm^{2}$ $\displaystyle 17.85\; cm^{2}$

Correct answer:

$17.85\; cm^{2}$ $\displaystyle 17.85\; cm^{2}$

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 $A=\frac{1}{2}base\cdot height$ $\displaystyle A=\frac{1}{2}base\cdot height$

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

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

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

$height=2.55\; cm$ $\displaystyle height=2.55\; cm$

Then we plug in all values for the equation:

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

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Example Question #12 : Solve A Right Triangle

Find the area of the given isosceles triangle:

Varsity log graph

Possible Answers:

$6.25 \: cm^{2}$ $\displaystyle 6.25 \: cm^{2}$

$3.11 \: cm^{2}$ $\displaystyle 3.11 \: cm^{2}$

$25.55\: cm^{2}$ $\displaystyle 25.55\: cm^{2}$

$3.34 \: cm^{2}$ $\displaystyle 3.34 \: cm^{2}$

$6.71 \: cm^{2}$ $\displaystyle 6.71 \: cm^{2}$

Correct answer:

$6.25 \: cm^{2}$ $\displaystyle 6.25 \: cm^{2}$

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:

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

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

$height=4.83\: cm$ $\displaystyle height=4.83\: cm$

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

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

$Base=2.59\: cm$ $\displaystyle Base=2.59\: cm$

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

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

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Example Question #13 : Solve A Right Triangle

Pcq1

Solve the right triangle given that a=5, b=12, and A=22.620°

Possible Answers:

B=67.380°

C=90°

c=13

B=67.380°

C=90°

c=17

B=90°

C=67.380°

c=13

B=90°

C=67.380°

c=169

None of these answers are correct.

Correct answer:

B=67.380°

C=90°

c=13

Explanation:

Pcq1

C is given as 90°.

A is given as 22.620°

a is given as 5

b is given as 12

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

Therefore...

$5^{2}+12^{2}=c^{2}$ $\displaystyle 5^{2}+12^{2}=c^{2}$

$169=c^{2}$ $\displaystyle 169=c^{2}$

$\sqrt{169}=\sqrt{c^{2}}$ $\displaystyle \sqrt{169}=\sqrt{c^{2}}$

13=c $\displaystyle 13=c$

All angles of a triangle add up to equal 180°.

$90^{\circ}+22.620^{\circ}+B=180^{\circ}$ $\displaystyle 90^{\circ}+22.620^{\circ}+B=180^{\circ}$

$112.620^{\circ}+B=180^{\circ}$ $\displaystyle 112.620^{\circ}+B=180^{\circ}$

$B=180^{\circ}-112.620^{\circ}$ $\displaystyle B=180^{\circ}-112.620^{\circ}$

$B=67.380^{\circ}$ $\displaystyle B=67.380^{\circ}$

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Precalculus : Solve a Right Triangle

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #11 : Solve A Right Triangle

Example Question #12 : Solve A Right Triangle

Example Question #13 : Solve A Right Triangle

All Precalculus Resources

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