write the electron dot structure of sodium cholride and formation of ionic bond
Answers
Explanation:
Given:
BC= 15cm
\rm{sin\ B=\dfrac{4}{5}}sin B=
5
4
To Find:
i) Measurement of AB and AC
ii) Measurement of CD and AD, if tan(∠ADC)=1
To Prove:
\longrightarrow\bf{tan^{2}\ B-\dfrac{1}{cos^{2}\ B}=-1}⟶tan
2
B−
cos
2
B
1
=−1
Solution:
In ΔABC,
\longrightarrow\rm{cos\ B=\sqrt{1-sin^{2}B}}⟶cos B=
1−sin
2
B
\longrightarrow\rm{cos\ B=\sqrt{1-\bigg(\dfrac{4}{5}\bigg)^{2}}}⟶cos B=
1−(
5
4
)
2
\longrightarrow\rm{cos\ B=\sqrt{1-\dfrac{16}{25}}}⟶cos B=
1−
25
16
\longrightarrow\rm{cos\ B=\sqrt{\dfrac{25-16}{25}}}⟶cos B=
25
25−16
\longrightarrow\rm{cos\ B=\sqrt{\dfrac{9}{25}}}⟶cos B=
25
9
\longrightarrow\rm{cos\ B=\dfrac{3}{5}}⟶cos B=
5
3
Now,
\longrightarrow\rm{cos\ B=\dfrac{BC}{AB}}⟶cos B=
AB
BC
\longrightarrow\rm{\dfrac{3}{5}=\dfrac{15}{AB}}⟶
5
3
=
AB
15
\longrightarrow\rm{AB=\dfrac{15\times5}{3}}⟶AB=
3
15×5
\longrightarrow\rm\green{AB=25\ cm}⟶AB=25 cm
Now,
\longrightarrow\rm{sin\ B=\dfrac{AC}{AB}}⟶sin B=
AB
AC
\longrightarrow\rm{\dfrac{4}{5}=\dfrac{AC}{25}}⟶
5
4
=
25
AC
\longrightarrow\rm{\dfrac{4\times25}{5}=AC}⟶
5
4×25
=AC
\longrightarrow\rm{AC=\dfrac{4\times25}{5}}⟶AC=
5
4×25
\longrightarrow\rm\green{AC=20\ cm}⟶AC=20 cm
━━━━━━━━━━━━━━━━━━━━━
Now, In ΔACD
tan(∠ADC)=1
That means,
∠ADC= 45°
Or
D= 45°
So,
\longrightarrow\rm{tan\ D=\dfrac{AC}{CD}}⟶tan D=
CD
AC
\longrightarrow\rm{tan\ 45^{\circ}=\dfrac{20}{CD}}⟶tan 45
∘
=
CD
20
\longrightarrow\rm{1=\dfrac{20}{CD}}⟶1=
CD
20
\longrightarrow\rm\green{CD=20\ cm}⟶CD=20 cm
Now,
\longrightarrow\rm{sin\ D=\dfrac{AC}{AD}}⟶sin D=
AD
AC
\longrightarrow\rm{sin\ 45^{\circ}=\dfrac{20}{AD}}⟶sin 45
∘
=
AD
20
\longrightarrow\rm{\dfrac{1}{\sqrt{2}} =\dfrac{20}{AD}}⟶
2
1
=
AD
20
\longrightarrow\rm\green{AD=20\sqrt{2}\ cm}⟶AD=20
2
cm
━━━━━━━━━━━━━━━━━━━━━
Now, we know that,
\pink{\underline{\boxed{\bf{1+tan^{2}\theta=sec^{2}\theta}}}}
1+tan
2
θ=sec
2
θ
\purple{\underline{\boxed{\bf{sec \theta}=\dfrac{1}{cos\theta}}}}
secθ=
cosθ
1
So, by using above properties
\longrightarrow\rm{1+tan^{2}B=sec^{2}B}⟶1+tan
2
B=sec
2
B
\longrightarrow\rm{tan^{2}B-sec^{2}B=-1}⟶tan
2
B−sec
2
B=−1
\longrightarrow\rm{tan^{2}B-\bigg(\dfrac{1}{cosB} \bigg)^{2}=-1}⟶tan
2
B−(
cosB
1
)
2
=−1
\longrightarrow\rm{tan^{2}B-\dfrac{1}{cos^{2}B}=-1}⟶tan
2
B−
cos
2
B
1
=−1
\bigstar\bf\red{Hence\ Proved}★Hence Proved
Hope it's help you...✌️❣️
☞ By losing an electron, Na becomes Na+ (cation) and this electron is accepted by Cl to form Cl- (anion). The oppositely charged ions attract each other and form NaCl. They are held together by strong electrostatic forces of attraction. ☜