Question

How to do this sum? Which have a triangle with AB=5cm and angle C with 30 degree

Answer 1

Answer:

answer for the given problem is given

Answer 2

$\huge\sf\pink{Answer}$

☞ AC is 10 cm & BC is 5√3 cm

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ AB = 5 cm

✭ ∠ACB = 30°

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ The length of AC & BC?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

We know that,

$\underline{\boxed{\sf sin\theta= \dfrac{Opposite}{Hypothenuse}}}$

Here for θ = 30°

◕ Opposite side = AB

◕ Adjacent side = BC

◕ Hypothenuse = AC

Substituting the given values,

➳ $\sf sin \ 30^{\circ} = \dfrac{5}{AC}$

$\bigg\lgroup\sf sin \ 30^{\circ} = \dfrac{1}{2}\bigg\rgroup$

➳ $\sf \dfrac{1}{2} = \dfrac{5}{AC}$

➳ $\sf AC = 5 \times 2$ «« Cross Multiply »»

➳ $\sf \red{AC = 10 \ cm}$

Similarly we know that,

$\underline{\boxed{\sf tan\theta = \dfrac{Opposite}{Adjacent}}}$

Substituting the values,

➝ $\sf tan \ 30^{\circ} = \dfrac{5}{BC}$

$\bigg\lgroup\sf tan \ 30^{\circ} = \dfrac{1}{\sqrt{3}}\bigg\rgroup$

➝ $\sf \dfrac{1}{\sqrt{3}} = \dfrac{5}{BC}$

➝ $\sf BC = 5 \times \sqrt{3}$ «« Cross Multiply »»

➝ $\sf \orange{BC = 5\sqrt{3} \ cm}$

━━━━━━━━━━━

More Trigonometric Values

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

━━━━━━━━━━━━━━━━━━