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Zain is vacationing in Italy. They were in Pisa to see the famous Leaning Tower when a question came across their mind. What would the tower's height be if it was not a leaning tower? Zain's distance to the tower is $80$ meters and they can measure an angle of elevation of $37^{\circ }$ to the tower's top. Furthermore, the guidebook states that the tower's inclination is about $4^{\circ }.$

Remembering a Geometry lesson, they realize that the situation can be modeled using a non-right triangle. Help Zain calculate the height of the upright tower. Write the answer rounded to one decimal place.

Solution

The situation can be modeled using a non-right triangle.

The tower's inclination angle and $\angle B$ are complementary angles, so the sum of their measures adds to $90^{\circ }.$ The measure of $\angle B$ can now be calculated.

$$\begin{array}{c}{4}^{\circ }+m\angle B=90^{\circ }\\ \Updownarrow \\ m\angle B=86^{\circ }\end{array}$$

Now that two angles and their included side are known, it is possible to use the Law of Sines. To do so, the measure of $\angle A$ must be found. This will be done by using the Triangle Angle Sum Theorem.

$$\begin{array}{c}m\angle A+86^{\circ }+37^{\circ }=180^{\circ }\\ \Updownarrow \\ m\angle A=57^{\circ }\end{array}$$

Finally, the Law of Sines will be used to write a proportion. This equation will be solved for $c,$ the height of the upright tower.

$\frac{a}{\sin A}=\frac{c}{\sin C}$

SubstituteValues

Substitute values

$\frac{80}{\sin 57^{\circ }}=\frac{c}{\sin 37^{\circ }}$

Solve for $c$

MultEqn

$\text{LHS}\cdot \sin 37^{\circ }=\text{RHS}\cdot \sin 37^{\circ }$

$\frac{80}{\sin 57^{\circ }}(\sin 37^{\circ })=c$

UseCalc

Use a calculator

$57.406571\dots =c$

RearrangeEqn

Rearrange equation

$c=57.406571\dots$

RoundDec

Round to $1$ decimal place(s)

$c\approx 57.4$

Therefore, the height of the Leaning Pisa Tower would be around $57.4$ meters if it was standing straight.

Ignacio's grandparent wants to construct a fence for a quadrilateral piece of land. To find the perimeter of the land, he starts measuring its sides using an old trundle wheel. Unfortunately, after measuring just two sides, the trundle wheel breaks.

Ignacio wants to help his grandparent and, in an attempt to simplify the problem, he divides the land into two triangles. Then, by using a compass, he is able to measure the angles of these triangles.

Solution

Note that two side lengths and all the angle measurements are known in $△ABC.$

In particular, since two side lengths and the measure of their included angle are known, the Law of Cosines can be used to solve for the missing side length.

$b^2=a^2+c^2-2ac\cos B$

SubstituteValues

Substitute values

$b^2=400^2+{560}^2-2(400)(560)\cos 80^{\circ }$

Solve for $b$

CalcPow

Calculate power

$b^2=160000+313600-2(400)(560)\cos 80^{\circ }$

AddTerms

Add terms

$b^2=473600-2(400)(560)\cos 80^{\circ }$

Multiply

Multiply

$b^2=473600-448000\cos 80^{\circ }$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$b=\sqrt{473600-448000\cos 80^{\circ }}$

UseCalc

Use a calculator

$b=629.130842\dots$

RoundInt

Round to nearest integer

$b\approx 629$

When solving the above equation, only the principal root was considered because side lengths are always positive. Therefore, the length of $\overline{AC}$ is about $629$ meters.

Now, $△ACD$ will be considered.

Since one side length and all the angle measures are known, the missing side lengths can be found by using the Law of Sines. The length of $\overline{DC}$ will be calculated.

$\frac{a}{\sin A}=\frac{d}{\sin D}$

SubstituteValues

Substitute values

$\frac{a}{\sin 21^{\circ }}=\frac{629}{\sin 120^{\circ }}$

Solve for $a$

MultEqn

$\text{LHS}\cdot \sin 21^{\circ }=\text{RHS}\cdot \sin 21^{\circ }$

$a=\frac{629}{\sin 120^{\circ }}(\sin 21^{\circ })$

UseCalc

Use a calculator

$a=260.285020\dots$

RoundInt

Round to nearest integer

$a\approx 260$

Therefore, the length of $\overline{DC}$ is about $260$ meters. Next, the length of $\overline{AD}$ can be calculated by following the same procedure.

$\frac{c}{\sin C}=\frac{d}{\sin D}$

SubstituteValues

Substitute values

$\frac{c}{\sin 39^{\circ }}=\frac{629}{\sin 120^{\circ }}$

Solve for $c$

MultEqn

$\text{LHS}\cdot \sin 39^{\circ }=\text{RHS}\cdot \sin 39^{\circ }$

$c=\frac{629}{\sin 120^{\circ }}(\sin 39^{\circ })$

UseCalc

Use a calculator

$c=457.079577\dots$

RoundInt

Round to nearest integer

$c\approx 457$

Now, all the sides of the piece of land are known.

Finally, the perimeter will be calculated by adding all the side lengths.

$$\begin{array}{c}P=457\text{ m}+560\text{ m}+400\text{ m}+260\text{ m}\\ \Updownarrow \\ P=1677\text{ m}\end{array}$$

The perimeter of the piece of land is about $1677$ meters.

The challenge presented at the beginning of this lesson can be solved by using a combination of the Law of Sines, the Law of Cosines, and the extended form of the Law of Sines.

The question here was to find the area of the circle. This will be answered by first finding the ratio of the triangle's side lengths to the sine of their opposite angles.

Solution

Since two side lengths and the included angle are known, the Law of Cosines can be used to find the missing side length. Because the missing side is opposite to the known angle, finding the side length will allow to calculate the desired ratio.

The Law of Cosines will be used to find the missing side $a.$

$a^2=b^2+c^2-2bc\cos A$

SubstituteValues

Substitute values

$a^2={4.9}^2+{4.6}^2-2(4.9)(4.6)\cos 44^{\circ }$

Solve for $a$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$a=\sqrt{4.9^2+4.6^2-2(4.9)(4.6)\cos 44^{\circ }}$

CalcPow

Calculate power

$a=\sqrt{24.01+21.16-2(4.9)(4.6)\cos 44^{\circ }}$

AddTerms

Add terms

$a=\sqrt{45.17-2(4.9)(4.6)\cos 44^{\circ }}$

Multiply

Multiply

$a=\sqrt{45.17-45.08\cos 44^{\circ }}$

UseCalc

Use a calculator

$a=3.569616\dots$

RoundDec

Round to $1$ decimal place(s)

$a\approx 3.6$

When solving the above equation, only the principal root was considered. This is because a side length is always positive. Therefore, the missing side length is about $3.6$ units.

Now that the measure of $\angle A$ and the length of its opposite side $a$ are known, the desired ratio can be calculated.

$\frac{a}{\sin A}$

SubstituteII

$a=3.6$, $A=44^{\circ }$

$\frac{3.6}{\sin 44^{\circ }}$

Evaluate

UseCalc

Use a calculator

$5.182403\dots$

RoundDec

Round to $1$ decimal place(s)

$5.2$

The ratio of a side length to the sine of its opposite angle is about $5.2.$ Therefore, by the extended form of the Law of Sines, the length of the circle's diameter is about $5.2.$ Recall that the radius of a circle is half its diameter.

$$\begin{array}{c}\text{Radius:}\frac{5.2}{2}=2.6\text{ units}\end{array}$$

With this information, the area of the circle can be calculated.

$A=\pi r^2$

Substitute

$r=2.6$

$A=\pi {\left(2.6\right)}^2$

Simplify right-hand side

CalcPow

Calculate power

$A=\pi (6.76)$

CommutativePropMult

\CommutativePropMult

$A=6.76\pi$

UseCalc

Use a calculator

$A=21.237166\dots$

RoundDec

Round to $1$ decimal place(s)

$A\approx 21.2$

The area of the circumscribed circle is about $21.2$ square units.