Mathematics, 13.06.2021 14:00 wilneishawatkins

5. Two obeserves Pand Q 15 m apart observe a kite in the same vertical plane and from the same side of
the kite. The angle of elevation of the kite from P and Q are
35° and 45°respectively. Find the height
ar the kite to the nearest metre.
(WAEC

Answers: 3

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Answer from: lberries08

Approximately $35\; \rm m$ .

Step-by-step explanation:

In a right triangle, the tangent of an angle is the ratio between the length of the side opposite to this angle and the length of the side adjacent to this angle:

$\displaystyle \tan{\theta} = \frac{\text{opposite}}{\text{adjacent}}$ .

Let denote the height of this kite in meters.

Refer to the diagram attached. Let $\sf O$ denote the point right below the kite in the same vertical plane as $\sf P$ and $\sf Q$ . Denote the kite as $\sf K$ .

The kite $\sf K$ , observer $\sf Q$ , and point $\sf O$ are the three vertices of right triangle $\triangle {\sf KQO}$ , where $\angle{\sf KQO} = 45^\circ$ .

In this right triangle, $\sf OK$ is the side opposite to $\angle{\sf KQO}$ , whereas $\sf OQ$ is the side adjacent to $\angle{\sf KQO}\!$ .

Therefore:

$\displaystyle \tan(\angle{\sf KQO}) = \frac{\text{length of ${\sf OK}$}}{\text{length of ${\sf OQ}$}}$ .

The length of $\sf OK$ is the same as the height of this kite,

meters.The length of $\sf OQ$ denotes the horizontal distance between the kite and observer $\sf Q$ .

$\displaystyle \tan(45^\circ) = \frac{x}{\text{length of ${\sf OQ}$}}$ .

$\text{length of ${\sf OQ}$} = x\, \tan(45^\circ)$ .

Similarly, in right triangle $\triangle {\sf KPO}$ , $\angle{\sf KPO} = 35^\circ$ . In this right triangle, $\sf OK$ is the side opposite to $\angle{\sf KPO}$ , whereas $\sf OP$ is the side adjacent to $\angle{\sf KPO}\!$ .

$\displaystyle \tan(\angle{\sf KPO}) = \frac{\text{length of ${\sf OK}$}}{\text{length of ${\sf OP}$}}$ .

$\displaystyle \tan(35^\circ) = \frac{x}{\text{length of ${\sf OP}$}}$ .

$\text{length of ${\sf OP}$} = x\, \tan(35^\circ)$

The question states that observer $\sf P$ and observer $\sf Q$ are on the same side of the kite. Hence, the horizontal distance between $\sf P\!$ and $\sf Q\!$ would be the same as the difference between:

the horizontal distance between $\sf P$ and the kite (same as the length of segment $\sf OP$ ,) andthe horizontal distance between $\sf Q$ and the kite (same as the length of segment $\sf OQ$ .)

In other words:

$\begin{aligned}& \text{horizontal distance between ${\sf P}$ and ${\sf Q}$} \\ &= (\text{horizontal distance between ${\sf P}$ and ${\sf K}$} \\ &\quad\quad -\text{horizontal distance between ${\sf Q}$ and ${\sf K}$})\end{aligned}$ .

$\begin{aligned}& \text{length of ${\sf PQ}$} \\ &= \text{length of ${\sf OP}$} - \text{length of ${\sf OQ}$}\end{aligned}$ .

$\displaystyle 15 = \frac{x}{\tan(35^\circ)} - \frac{x}{\tan(45^\circ)}$ .

Solve for :

$\begin{aligned}x &= 15 \left/\left(\frac{1}{\tan({35}^\circ)} - \frac{1}{\tan({45}^\circ)}\right)\right. \approx 35\end{aligned}$ .

5. Two obeserves Pand Q 15 m apart observe a kite in

the same vertical plane and from the same side

Answer from: Quest

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

Which expression is the simplest form of -(4x + y) + 2(x – 3y)?

Answer from: Quest

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

happy to

pls mark as brainliest

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