We are given two problems. (a) A cylinder with radius $3.5$ cm has its two ends closed. If the total surface area is $209 \text{ cm}^2$, calculate the height of the cylinder. Take $\pi = \frac{22}{7}$. (b) In a circle with center $O$, $ABC$ is a tangent at $B$. Given $\angle BDF = 66^\circ$ and $\angle DBC = 57^\circ$, calculate (i) $\angle EBF$ and (ii) $\angle BGF$.

1. Problem Description

We are given two problems.

(a) A cylinder with radius $3.5$ cm has its two ends closed. If the total surface area is $209 \text{ cm}^2$, calculate the height of the cylinder. Take $\pi = \frac{22}{7}$.

(b) In a circle with center $O$, $ABC$ is a tangent at $B$. Given $\angle BDF = 66^\circ$ and $\angle DBC = 57^\circ$, calculate (i) $\angle EBF$ and (ii) $\angle BGF$.

2. Solution Steps

(a)

The total surface area of a cylinder with radius $r$ and height $h$ is given by

$A = 2\pi r^2 + 2\pi rh$.

We are given $r = 3.5$ cm and $A = 209 \text{ cm}^2$. Also, $\pi = \frac{22}{7}$.

Substituting these values into the formula:

$209 = 2 \times \frac{22}{7} \times (3.5)^2 + 2 \times \frac{22}{7} \times 3.5 \times h$

$209 = 2 \times \frac{22}{7} \times 12.25 + 2 \times \frac{22}{7} \times 3.5 \times h$

$209 = 2 \times 22 \times 1.75 + 2 \times 22 \times 0.5 \times h$

$209 = 44 \times 1.75 + 22h$

$209 = 77 + 22h$

$22h = 209 - 77$

$22h = 132$

$h = \frac{132}{22}$

$h = 6$ cm

(b)

(i) We are given that $ABC$ is a tangent to the circle at $B$, and $\angle DBC = 57^\circ$.

By the alternate segment theorem, $\angle DEB = \angle DBC = 57^\circ$.

Since $E$ is on the circle, $\angle EDB$ and $\angle EFB$ subtend the same arc, so $\angle EFB = \angle EDB$. Also, $\angle BDF = 66^\circ$, so $\angle EDB = 180^\circ - 66^\circ - 66^\circ = \angle EDB = 66^\circ$.

Since angles in the same segment are equal, $\angle BEF = \angle BDF = 66^\circ$.

$\angle EBF = 180^\circ - \angle BEF - \angle EFB = 180^\circ - \angle DEB - \angle EDB = 180^\circ - 57^{\circ} - 66^{\circ} = 180^{\circ} -123^{\circ}=57^{\circ}$.

$\angle EBF = 57^\circ$

(ii) $\angle EDF = 66^{\circ}$. Since the angle at the centre is twice the angle at the circumference,

$\angle EBF = \angle DBC = 57^{\circ}$. Also, the sum of the angles in triangle $EBF$ is $180^\circ$.

Since $EB$ is a diameter, $\angle EFB = 90^\circ$, and $EBF+BEF=90^\circ$.

Since angle in the same segment are equal, $\angle BEF = \angle BDF = 66^\circ$.

$\angle EBF = 180^\circ - (66^\circ + 57^\circ) = 57^\circ$. $\angle EBF = 180^\circ - 90^\circ = 90^\circ$.

Since $EB$ is a diameter, $\angle EDB=90^\circ$.

$\angle FBE = 90^\circ-66^\circ=24^\circ$. $EBD = 57$.

$\angle EBD + \angle DBC = 57 = 57$ and $EBD=33^\circ$.

$\angle EBF$ is equal to $90$.

$\angle BGF$ is at the centre so is twice the angle $BDF$.

$\angle BGF=2\angle BEF$ and $EF=90^\circ$.

$\angle BEF + \angle FED = 90^{\circ}$

$\angle BGF = 2 \times \angle BDF$. $\angle BDF=66^\circ$.

$\angle BGF = 132^{\circ}$.

3. Final Answer

(a) The height of the cylinder is $6$ cm.

(b) (i) $\angle EBF = 57^\circ$

(ii) $\angle BGF = 132^\circ$