Question

1. a man flies a kite with a 120 - foot string. the angle the string forms with the ground is 61°. how high off the ground is the kite?
2. clara is standing at the base of a 25 - foot ladder that she is using to scale a tree. if the base of the ladder is 13 feet from the tree, at what angle of elevation is the ladder?
3. from the top of a lighthouse 210 feet high, the angle of depression of a boat is 27°. find the distance from the boat to the foot of the lighthouse. the lighthouse was built at sea level.

bonus in an isosceles triangle, the base is 10 cm long, and the legs are 8 cm long. find the measure of the vertex ∠bac (hint: d is a perpendicular bisector of bc)

Explanation:

Step1: Solve for height of kite

We have a right - triangle where the string of the kite is the hypotenuse ($c = 120$ feet) and we want to find the height ($h$) which is the side opposite the given angle $\theta=61^{\circ}$. Using the sine function $\sin\theta=\frac{opposite}{hypotenuse}$. So, $h = c\times\sin\theta$.
$h = 120\times\sin(61^{\circ})$
Since $\sin(61^{\circ})\approx0.8746$, then $h\approx120\times0.8746 = 104.952\approx105$ feet.

Step2: Solve for angle of elevation of the ladder

In a right - triangle with ladder length (hypotenuse) $c = 25$ feet and base $a = 13$ feet. We use the cosine function $\cos\theta=\frac{adjacent}{hypotenuse}$. So, $\cos\theta=\frac{13}{25}=0.52$. Then $\theta=\cos^{- 1}(0.52)\approx58.78^{\circ}\approx59^{\circ}$.

Step3: Solve for distance from boat to lighthouse

The angle of depression from the top of the lighthouse to the boat is $27^{\circ}$. The height of the lighthouse (opposite side) is $h = 210$ feet. We want to find the distance $d$ (adjacent side) from the boat to the lighthouse. Using the tangent function $\tan\theta=\frac{opposite}{adjacent}$. So, $\tan(27^{\circ})=\frac{210}{d}$, then $d=\frac{210}{\tan(27^{\circ})}$. Since $\tan(27^{\circ})\approx0.5095$, then $d=\frac{210}{0.5095}\approx412$ feet.

Step4: Solve for angle in isosceles triangle

Let the isosceles triangle have base $BC = 10$ cm and legs $AB = AC=8$ cm. Let $AD$ be the perpendicular bisector of $BC$, so $BD=\frac{BC}{2}=5$ cm. Using the cosine function in right - triangle $ABD$, $\cos\angle BAD=\frac{AD}{AB}$. First, find $AD$ using the Pythagorean theorem in $\triangle ABD$: $AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{8^{2}-5^{2}}=\sqrt{64 - 25}=\sqrt{39}\approx6.245$ cm. Then $\cos\angle BAD=\frac{5}{8}=0.625$, and $\angle BAD=\cos^{-1}(0.625)\approx51.32^{\circ}$. So, $\angle BAC = 2\angle BAD\approx102.64^{\circ}\approx103^{\circ}$.

Answer:

Height of Kite: 105 feet
Angle of Elevation: $59^{\circ}$
Distance: 412 feet
BONUS: $103^{\circ}$