Answer
Verified
465k+ views
Hint: As, in the $\Delta ABC$ , we are given with angle A and length of the side AB. So, now we can use simple trigonometric ratios of sin and cosine to find the length of the remaining two sides and the angle B of the triangle ABC.
Complete step-by-step answer:
In the above $\Delta ABC$ , concerning the $\angle A$ side BC is the perpendicular, side AC is base and side AB is the hypotenuse respectively.
We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{BC}{AB}$ .
The angle that we are considering is $\angle A$ . So , $\theta =\angle A=30{}^\circ $ .
Also, we know that $sin30{}^\circ =\dfrac{1}{2}$ .
$\Rightarrow sin\theta =sin30{}^\circ =\dfrac{1}{2}=\dfrac{BC}{AB}.........(i)$
Now, in the question, we are given that the length of side AB is given as AB = 40 units . So, substituting AB = 40 units in equation ( i ), we get:
$\begin{align}
& \dfrac{BC}{AB}=\dfrac{1}{2} \\
& \Rightarrow \dfrac{BC}{40}=\dfrac{1}{2} \\
& \Rightarrow BC=\dfrac{40}{2} \\
& \Rightarrow BC=20 \\
\end{align}$
$\Rightarrow $ Value of BC = 20 units.
Now, we know that in a right-angled triangle, if we know the length of any two sides of the triangle, we can find the third side using the Pythagorean Theorem.
Pythagoras theorem says that in a right-angled triangle, $Hypotenus{{e}^{2}}=Bas{{e}^{2}}+Perpendicula{{r}^{2}}$
Here, the unknown side is AC. So, according to Pythagoras Theorem, $A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$
Hence, by substituting the values of AB and BC, we get:
${{40}^{2}}=A{{C}^{2}}+{{20}^{2}}$
$\Rightarrow A{{C}^{2}}={{40}^{2}}-{{20}^{2}}$
$\Rightarrow A{{C}^{2}}=1600-400$
$\begin{align}
& \Rightarrow A{{C}^{2}}=1200 \\
& \Rightarrow AC=\sqrt{1200} \\
& \Rightarrow AC=20\sqrt{3} \\
\end{align}$
Therefore, the value of $AC=20\sqrt{3}$units.
Now, we have to find the angle B of the triangle ABC. Let us assume that the angle B = $\theta $ , i.e. $\angle B=\theta $ .
Now, in the above $\Delta ABC$ , concerning the $\angle B$ side AC is the perpendicular, side BC is base and side AB is the hypotenuse respectively.
We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{AC}{AB}$ .
Here, $AC=20\sqrt{3}$ and AB = 40 units.
$\begin{align}
& \Rightarrow \sin \theta =\dfrac{20\sqrt{3}}{40}=\dfrac{\sqrt{3}}{2} \\
& \Rightarrow \sin \theta =\dfrac{\sqrt{3}}{2} \\
& \Rightarrow \theta =60{}^\circ \\
\end{align}$
Hence, the value of $\angle B=60{}^\circ $ .
Note: Be careful while substituting the values of the trigonometric functions. Students generally get confused and make silly mistakes.
Complete step-by-step answer:
In the above $\Delta ABC$ , concerning the $\angle A$ side BC is the perpendicular, side AC is base and side AB is the hypotenuse respectively.
We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{BC}{AB}$ .
The angle that we are considering is $\angle A$ . So , $\theta =\angle A=30{}^\circ $ .
Also, we know that $sin30{}^\circ =\dfrac{1}{2}$ .
$\Rightarrow sin\theta =sin30{}^\circ =\dfrac{1}{2}=\dfrac{BC}{AB}.........(i)$
Now, in the question, we are given that the length of side AB is given as AB = 40 units . So, substituting AB = 40 units in equation ( i ), we get:
$\begin{align}
& \dfrac{BC}{AB}=\dfrac{1}{2} \\
& \Rightarrow \dfrac{BC}{40}=\dfrac{1}{2} \\
& \Rightarrow BC=\dfrac{40}{2} \\
& \Rightarrow BC=20 \\
\end{align}$
$\Rightarrow $ Value of BC = 20 units.
Now, we know that in a right-angled triangle, if we know the length of any two sides of the triangle, we can find the third side using the Pythagorean Theorem.
Pythagoras theorem says that in a right-angled triangle, $Hypotenus{{e}^{2}}=Bas{{e}^{2}}+Perpendicula{{r}^{2}}$
Here, the unknown side is AC. So, according to Pythagoras Theorem, $A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$
Hence, by substituting the values of AB and BC, we get:
${{40}^{2}}=A{{C}^{2}}+{{20}^{2}}$
$\Rightarrow A{{C}^{2}}={{40}^{2}}-{{20}^{2}}$
$\Rightarrow A{{C}^{2}}=1600-400$
$\begin{align}
& \Rightarrow A{{C}^{2}}=1200 \\
& \Rightarrow AC=\sqrt{1200} \\
& \Rightarrow AC=20\sqrt{3} \\
\end{align}$
Therefore, the value of $AC=20\sqrt{3}$units.
Now, we have to find the angle B of the triangle ABC. Let us assume that the angle B = $\theta $ , i.e. $\angle B=\theta $ .
Now, in the above $\Delta ABC$ , concerning the $\angle B$ side AC is the perpendicular, side BC is base and side AB is the hypotenuse respectively.
We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{AC}{AB}$ .
Here, $AC=20\sqrt{3}$ and AB = 40 units.
$\begin{align}
& \Rightarrow \sin \theta =\dfrac{20\sqrt{3}}{40}=\dfrac{\sqrt{3}}{2} \\
& \Rightarrow \sin \theta =\dfrac{\sqrt{3}}{2} \\
& \Rightarrow \theta =60{}^\circ \\
\end{align}$
Hence, the value of $\angle B=60{}^\circ $ .
Note: Be careful while substituting the values of the trigonometric functions. Students generally get confused and make silly mistakes.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
A group of fish is known as class 7 english CBSE
The highest dam in India is A Bhakra dam B Tehri dam class 10 social science CBSE
Write all prime numbers between 80 and 100 class 8 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Onam is the main festival of which state A Karnataka class 7 social science CBSE
Who administers the oath of office to the President class 10 social science CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Kolkata port is situated on the banks of river A Ganga class 9 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE