Question

A. For the point (3, 10), the distance from the y-axi is ------- units​

Answer 1

Answer:

\begin{gathered}{\bf{Given\::}} \\ \end{gathered}

Given:

The angle of elevation of the top of a building from foot of the tower is 30°.

The angle of elevation of the top of the tower from the root of the building is 45°.

Height of the tower is 30 m.

\begin{gathered} \\ {\bf{To\: Find\::}} \\ \end{gathered}

ToFind:

The height of the building.

\begin{gathered} \\ {\bf{Solution\::}} \\ \end{gathered}

Solution:

Let,

As shown in figure,

Height of the building = PQ.

Height of the tower = RS = 30 m.

Angle of elevation of the top of a building from foot of the tower = \rm{\angle{PRQ}}∠PRQ = 30°

Angle of elevation of the top of the tower from the root of the building = \rm{\angle{SQR}}∠SQR = 45°

Since,

Building & tower are perpendicular to the ground.

\longrightarrow\:\bf{\angle{PQR}\:=\:\angle{SRQ}\:=\:90^{\circ}\:}⟶∠PQR=∠SRQ=90

∘

Now,

In right angle ∆SQR,

➠ \tt{\tan{45^{\circ}}\:=\:\dfrac{SR}{QR}\:}tan45

∘

=

QR

SR

➠ \tt{1\:=\:\dfrac{30}{QR}\:}1=

QR

30

➠ \bf{QR\:=\:30\:cm}QR=30cm

Again,

In right angle ∆PQR,

➠ \begin{gathered}\tt{\tan{30^{\circ}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}

tan30

∘

=

30

PQ

➠ \begin{gathered}\tt{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}

3

1

=

30

PQ

➠ \tt{PQ\:=\:\dfrac{1}{\sqrt{3}}\times{30}\:}PQ=

3

1

×30

➠ \bf\pink{PQ\:=\:10\sqrt{3}\:cm\:=\:17.32\:cm}PQ=10

3

cm=17.32cm

∴ Height of the building is 17.32 cm.

Answer 2

Answer:

7

Step-by-step explanation: