1. The angle of elevation of a pastor from a member siting at the first floor in the church building is $$ 30^{\circ} $$. Given that the distance between the pastor and the member is 25 m and the height of the pastor can be ignored, find the vertical height between the foot of the pastor and the floor of the building. [Take $$ \sin (30)^{\circ}=\frac{1}{2} $$ ] 2. A farmer saw a deer on top of a hill. The foot of the farmer is on the same horizontal level as the base of the hill. If the angle of elevation of the farmer from the deer is $$ 50^{\circ} $$ and the distance between the farmer and the base of the hill is 20 m , $$ \left[ ight. $$ Take $$ an (50)=1.2^{\circ} $$ ] (a) represent the information on a diagram; (b) find the vertical height of the hill, ignore the height of the farmer. 3. A boy observes a friend on a storey building at an angle of elevation of $$ 30^{\circ} $$. If the distance between the boy and the foot of the building is 15 m , calculate the vertical height of the building. [Take $$ \sin (30)=0.8^{\circ} $$ ] 4. A hunter of height 1.7 m observed $$ a $$ bird on top of a tree at an angle of elevation of $$ 37^{\circ} $$. The distance between the foot of the hunter and the foot of the tree is 18 m . (a) Represent the information on a diagram. (b) Find the distance between the hunter and the bird. (c) Find the height of the tree. [Take $$ \sin (37)=0.76^{\circ} $$ ]

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1. The angle of elevation of a pastor from a member siting at the first floor in the church building is $$ 30^{\circ} $$. Given that the distance between the pastor and the member is 25 m and the height of the pastor can be ignored, find the vertical height between the foot of the pastor and the floor of the building. [Take $$ \sin (30)^{\circ}=\frac{1}{2} $$ ] 2. A farmer saw a deer on top of a hill. The foot of the farmer is on the same horizontal level as the base of the hill. If the angle of elevation of the farmer from the deer is $$ 50^{\circ} $$ and the distance between the farmer and the base of the hill is 20 m , $$ \left[ight. $$ Take $$ 	an (50)=1.2^{\circ} $$ ] (a) represent the information on a diagram; (b) find the vertical height of the hill, ignore the height of the farmer. 3. A boy observes a friend on a storey building at an angle of elevation of $$ 30^{\circ} $$. If the distance between the boy and the foot of the building is 15 m , calculate the vertical height of the building. [Take $$ \sin (30)=0.8^{\circ} $$ ] 4. A hunter of height 1.7 m observed $$ a $$ bird on top of a tree at an angle of elevation of $$ 37^{\circ} $$. The distance between the foot of the hunter and the foot of the tree is 18 m . (a) Represent the information on a diagram. (b) Find the distance between the hunter and the bird. (c) Find the height of the tree. [Take $$ \sin (37)=0.76^{\circ} $$ ]

UpStudy AI · Accepted Answer

1. We are told that a member on the first floor sees the pastor (whose own height is negligible) at an angle of elevation of
$$
30^\circ,
$$
and the distance (the line‐of‐sight, i.e. the hypotenuse of the right triangle) is 25 m. The vertical height, $$h$$, that we wish to find (the height difference between the member’s floor and the pastor’s foot) is given by
$$
h=25\sin(30^\circ).
$$
Using
$$
\sin(30^\circ)=\frac{1}{2},
$$
we have
$$
h=25\left(\frac{1}{2}\right)=12.5\text{ m}.
$$

2. (a) Make a diagram as follows. Draw a horizontal line representing ground level and mark a point $$F$$ for the farmer’s foot. At a distance of 20 m from $$F$$ along the horizontal line mark the base of the hill. Then, at the top of the hill mark the deer’s location (say point $$D$$). Join $$F$$ to $$D$$. The line $$FD$$ makes an angle of $$50^\circ$$ with the horizontal (this is the angle of elevation as seen from the deer that “sees” the farmer).  
 
(b) Let $$h$$ be the vertical height of the hill (that is, the vertical distance from the base of the hill to point $$D$$). In the right triangle with horizontal leg 20 m and vertical leg $$h$$, the tangent of the angle at $$F$$ is given by
$$
\tan(50^\circ)=\frac{h}{20}.
$$
We are given
$$
\tan(50^\circ)=1.2.
$$
Thus,
$$
h=20\times 1.2=24\text{ m}.
$$

3. A boy observes a friend on a storey building at an angle of elevation of
$$
30^\circ,
$$
and the horizontal distance from the boy to the foot of the building is 15 m. (A note appears to give $$\sin(30)=0.8$$; since the horizontal distance is given, it is more convenient to use the tangent function here. We then interpret the given value as the value of $$\tan(30^\circ)$$.) Let $$h$$ be the height of the building. Then
$$
\tan(30^\circ)=\frac{h}{15}.
$$
Taking
$$
\tan(30^\circ)=0.8,
$$
we deduce
$$
h=15\times0.8=12\text{ m}.
$$

4. (a) Draw the following diagram. Let point $$H$$ be the position at the foot of the hunter and point $$T$$ the base of the tree; the horizontal distance $$HT$$ is 18 m. The hunter, with height 1.7 m, is observing a bird at the top of the tree (point $$B$$). From the top of the hunter’s head, draw the line of sight to $$B$$ making an angle of $$37^\circ$$ with the horizontal.

(b) Let $$d$$ be the distance from the hunter’s head to the bird (i.e. the line‐of‐sight distance in the triangle with angle $$37^\circ$$ at the hunter’s head). Let $$x$$ be the vertical difference between the bird and the hunter’s head. Then by the definition of sine we have
$$
\sin(37^\circ)=\frac{x}{d}.
$$
That is,
$$
d=\frac{x}{\sin(37^\circ)}.
$$
We take
$$
\sin(37^\circ)=0.76.
$$

(c) Also note that in the right triangle formed by the horizontal distance (18 m) and the vertical difference $$x$$, the Pythagorean theorem yields
$$
d=\sqrt{18^2+x^2}.
$$
Equate the two expressions for $$d$$:
$$
\sqrt{18^2+x^2}=\frac{x}{0.76}.
$$
Square both sides:
$$
18^2+x^2=\frac{x^2}{0.5776}\quad\text{(since }0.76^2=0.5776\text{)}.
$$
Calculate $$18^2=324$$. Hence,
$$
324+x^2=\frac{x^2}{0.5776}.
$$
Multiply both sides by 0.5776:
$$
324\times0.5776+0.5776\,x^2=x^2.
$$
Compute:
$$
324\times0.5776=187.0224.
$$
Thus,
$$
187.0224+0.5776\, x^2=x^2.
$$
Subtract $$0.5776\, x^2$$ from both sides:
$$
187.0224= x^2-0.5776\, x^2=0.4224\, x^2.
$$
Thus,
$$
x^2=\frac{187.0224}{0.4224}\approx442.8,
$$
and
$$
x\approx\sqrt{442.8}\approx21.03\text{ m}.
$$
Now, the vertical height of the tree, $$H_{tree}$$, equals the hunter’s height plus $$x$$:
$$
H_{tree}=21.03+1.7\approx22.73\text{ m}.
$$
Also, now find the distance $$d$$ from the hunter’s head to the bird:
$$
d=\frac{21.03}{0.76}\approx27.67\text{ m}.
$$

Thus, the answers are:

1. $$12.5$$ m.
2. (a) [See diagram above]. (b) $$24$$ m.
3. $$12$$ m.
4. (a) [See diagram above]. (b) $$27.67$$ m (approximately). (c) $$22.73$$ m (approximately).
1. The vertical height between the pastor and the floor is 12.5 meters. 2. (a) [Diagram not provided]. (b) The hill is 24 meters tall. 3. The building is 12 meters high. 4. (a) [Diagram not provided]. (b) The distance from the hunter to the bird is approximately 27.67 meters. (c) The tree is approximately 22.73 meters tall.

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