Question

In the given figure, AE perpendicular BC, <B = 60°, <C = 30°, AB = 8 cm and BC = 24 cm, find:
(1) BE
(2) AC​

Answer 1

Trigonometric equation

The following are the tips and concept that can be use to find the solution:

• Having a basic knowledge of Trigonometric ratios and Angles.
• Trigonometric ratios are sin, cos, tan, cot, sec, cosec.
• The standard angles of these trigonometric ratios are 0°, 30°, 45°, 60° and 90°.
• Relationship between sides and T ratios

Analyse the values of important angles for all the six trigonometric ratios shown in the table given below:

$\boxed{\begin{array}{c |c|c|c|c|c} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}$

Relationship between sides and T ratios:

• sin(θ) = Height/Hypotenuse
• cos(θ) = Base/Hypotenuse
• tan(θ) = Height/Base
• cot(θ) = Base/Height
• sec(θ) = Hypotenuse/Base
• cosec(θ) = Hypotenuse/Height

Let's head to the Question now:

We are given that, In right ∆ABC, AE perpendicular to BC, ∠B = 60°, ∠C = 30°, AB = 8cm and BC = 24cm. With this information, we are asked to find out the value of (1) BE and (2) AC.

1. In right right ∆AEB, we have;

$\implies \sin( {60}^{\circ}) = \dfrac{AE}{AB} \\ \\ \implies \frac{ \sqrt{3} }{2} = \dfrac{AE}{8} \\ \\ \implies \cancel{8} \sqrt{3} = \cancel{2}AE \\ \\\implies AE = 4 \sqrt{3}$

Now, by using Pythagoras theorem;

$\implies (BE)^2 = (AB)^2 - (AE)^2 \\ \\ \implies BE = \sqrt{(AB)^2 - (AE)^2} \\ \\ \implies BE = \sqrt{ {(8)}^{2} - { \big(4 \sqrt{3} \big)}^{2} } \\ \\ \implies BE = \sqrt{64 - 48} \\ \\ \implies BE = \sqrt{16} \\ \\ \implies \boxed{\bf{BE =4}}$

Now finding the value of EC;

$\implies EC = BC - BE \\ \\ \implies EC = 24 - 4 \\ \\ \implies EC = 20$

2. In right ∆AEC and using Pythagoras theorem, we have;

$\implies (AC)^2 = (AE)^2 - (EC)^2 \\ \\\implies AC = \sqrt{(AE)^2 - (EC)^2} \\ \\ \implies AC = \sqrt{ \big(4 \sqrt{3} \big) ^{2} + {(20)}^{2} } \\ \\ \implies AC = \sqrt{48 + 400} \\ \\ \implies AC = \sqrt{448} \\ \\ \implies \boxed{\bf{AC =8 \sqrt{7}}}$

Hence, the value of BE and AC are 4cm and 8√7cm respectively.

Answer 2

Answer:

Question :-

✯ In the given figure, AE perpendicular BC, <B = 60°, <C = 30°, AB = 8 cm and BC = 24 cm, find :

☯ (1) BE

☯ (2) AC

Given :

✯ In the given figure, AE perpendicular BC, <B = 60°, <C = 30°, AB = 8 cm and BC = 24 cm.

Find Out :-

✯ Find :

☯ (1) BE

☯ (2) AC

Solution :-

✭ In case of AE :-

$\sf \longrightarrow \sin{60}^{\circ} = \dfrac{AE}{AB}$

$\sf \longrightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AE}{8}$

$\sf \longrightarrow {\small{\bold{\purple{\underline{AE = 4 \sqrt{3}\: cm}}}}}$

✭ In case of BE :-

$\sf \longrightarrow (BE)^2 = (AB)^2 - (AE)^2$

$\sf \longrightarrow (AB)^2 - (AE)^2$

$\sf \longrightarrow {(8)}^{2} - (4 \sqrt{3})^{2}$

$\sf \longrightarrow 64 - 48$

$\sf \longrightarrow 16$

$\longrightarrow \sf {\small{\bold{\purple{\underline{BE =4\: cm}}}}}$

✭ In case of EC :-

$\sf \longrightarrow EC = BC - BE$

$\sf \longrightarrow EC = 24 - 4$

$\longrightarrow \sf {\small{\bold{\purple{\underline{EC = 20\: cm}}}}}$

✭ In case of AC :

$\sf \longrightarrow (AC)^2 = (AE)^2 - (EC)^2$

$\sf \longrightarrow (AE)^2 - (EC)^2$

$\sf \longrightarrow (4 \sqrt{3} ) ^{2} + {(20)}^{2}$

$\sf \longrightarrow48 + 20$

$\sf \longrightarrow448$

$\longrightarrow \sf {\small{\bold{\purple{\underline{AC =8 \sqrt{7}\: cm}}}}}$

Henceforth, the value of BE and AC are 4 cm and 8√7 cm respectively.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

$\red{ \boxed{\sf{Trigonometric\: Ratios\: Table :-}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty\: \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered} \end{gathered}$

$\bigstar\: { \red{ \bf{ Information \: related \: to \:Trigonometry:}}}$

${ \green{ \bf{ sin θ = Perpendicular/Hypotenuse }}}$

${ \green{ \bf{ cos θ = Base/Hypotenuse }}}$

${ \green{ \bf{tan θ = Perpendicular/Base }}}$

${ \green{ \bf{sec θ = Hypotenuse/Base }}}$

${ \green{ \bf{ cosec θ = Hypotenuse/Perpendicular }}}$

${ \green{ \bf{ cot θ = Base/Perpendicular }}}$

$\bigstar { \red{ \bf{Their \: reciprocal \: Identities: }}}$

${ \green{ \bf{ cosec θ = \dfrac{1}{sin θ} }}}$

${ \green{ \bf{ sec θ = \dfrac{1}{cos θ} }}}$

${ \green{ \bf{ cot θ = \dfrac{1}{tan θ} }}}$

${ \green{ \bf{sin θ = \dfrac{1}{cosec θ} }}}$

${ \green{ \bf{ cos θ = \dfrac{1}{sec θ} }}}$

${ \green{ \bf{ tan θ = \dfrac{1}{cot θ}}}}$

$\bigstar { \red{ \bf{ Their \: fundamental \: trigonometric \: identities: }}}$

${ \green{ \bf{ sin^2θ + cos^2θ = 1 }}}$

${ \green{ \bf{ sec^2θ - tan^2θ = 1 }}}$

${ \green{ \bf{ cosec^2θ - cot^2θ = 1 }}}$