Question

Fig. 11.12
In Fig. 11.13, calculate a Ô AB
3com
B
18 cm​

Answer 1

Answer:

fig 11.12

in fig the 11.13 JCB thanks again time for a temporary one is needed and gone to go with further information or if your question you may

Answer 2

Correct Question: In fig. 11.13, AO = 3 cm and OT = 18 cm. calculate

a) $\hat O$, i.e., ∠AOT

b) $|AB|$

Answer:

a) $\hat O$ = 80.40°

b) $|AB|=\sqrt{\frac{35}{2} }$ cm

Step-by-step explanation:

a)

Since ΔAOT is a right angled triangle. So,

By Pythagoras theorem,

$OT^2=AO^2+AT^2$

$18^2=3^2+AT^2$

$324=9+AT^2$

$AT^2=324-9=315$

$AT=3\sqrt{35}$ cm

We know that,

$cos\ x=\frac{AO}{OT}$

$cos\ x=\frac{3}{18}$

$cos\ x=\frac{1}{6}$

     $x=cos^{-1}(\frac{1}{6} )$

     $x=80.40$ degree (Approx.)

Thus, $\hat O$ = 80.40°.

b)

From the figure, AB is perpendicular to OT.

So, ΔAOB and ΔABT are the right angled triangles.

Let $OB=y$ cm. Then, $BT=18-y$ cm

In ΔAOB, by Pythagoras theorem,

$AO^2=AB^2+OB^2$

$3^2=AB^2+y^2$ (Since $AO=3$ cm)

$AB^2=3^2-y^2$

$AB^2=9-y^2$. . . . . (1)

In ΔABT, by Pythagoras theorem,

$AT^2=AB^2+BT^2$

$({3\sqrt{35}})^2 =AB^2+{(18-y)}^2$ (Since $AT=3\sqrt{35}$ cm and $Bt=18-y$ cm)

$315=AB^2+324+y^2-36y$

$AB^2=315-324-y^2+36y$

$AB^2=-9-y^2+36y$ . . . . . (2)

From (1) and (2), we have

⇒ $9-y^2=-9-y^2+36y$

On simplifying we get,

⇒ $36y=18$

⇒ $y=\frac{18}{36}$

⇒ $y=1/2$

Substituting the value $y=1/2$ in the equation (1) as follows:

$AB^2=9-\left(\frac{1}{2} \right)^2$

$AB^2=9-\frac{1}{4}$

$AB^2=\frac{36-1}{2}$

$AB^2=\frac{35}{2}$

$AB=\sqrt{\frac{35}{2} }$ cm

⇒ $|AB|=\sqrt{\frac{35}{2} }$ cm

Therefore, the value of $|AB|$ is $\sqrt{\frac{35}{2} }$.

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