Question

In triangle ABC A=30 B=60 C=10 .find a and b

Answer 1

Answer:

f a triangle has angles of 30, 60 and 90 degrees

it is called one of the SPECIAL TRIANGLES.

It is used extensively in mathematics because we can find EXACT answers for sine, cosine and tangent of the angles 30 and 60 degrees.

All the following triangles have angles of 30, 60 and 90 degrees!

I think you will realise by now that these are just enlargements of each other.

They all have angles of 30, 60 and 90 degrees but the sides all have different lengths.

Fig 2 is the most common one. The triangles are special because we can just “read off” the values of sine, cosine and tangent of 30 and 60 degrees.

Step-by-step explanation:

Answer 2

There are some errors in the question. The correct question is read as:

In triangle $\[{\rm{ABC}}\]$, $\[{\rm{A = 30}}^\circ \]$, $\[{\rm{B = 60}}^\circ \]$ and $\[c = 10\]$. Find  $a$ and $b$.

Given:

In $\[\Delta \,{\rm{ABC}}\]$,

$\[{\rm{A = 30}}^\circ \]$, $\[{\rm{B = 60}}^\circ \]$ and $\[c = 10\]$

To Find:

Length of the sides $a$ and $b$

Solution:

We know that, the sum of all the angles of a triangle is equal to $% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later!% MathType!MTEF!2!1!+-% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaiI% dacaaIWaGaeyiSaalaaa!3A16!\[180^\circ \]% MathType!End!2!1!$.

Therefore,

$\[\begin{array}{l}{\rm{A + B + C = 180}}^\circ \\ \Rightarrow {\rm{C = 180}}^\circ - \left( {{\rm{A + B}}} \right)\\ \Rightarrow {\rm{C}} = {\rm{180}}^\circ - \left( {{\rm{30}}^\circ {\rm{ + 60}}^\circ } \right)\\ \Rightarrow {\rm{C = 180}}^\circ - 9{\rm{0}}^\circ \\ \Rightarrow {\rm{C = 90}}^\circ \end{array}\]$

Using sine rule:

$\[\frac{a}{{\sin {\rm{A}}}} = \frac{b}{{\sin {\rm{B}}}} = \frac{c}{{\sin {\rm{C}}}}\]$

Substituting the values in the above equation, we get:

$\[\frac{a}{{\sin 30^\circ }} = \frac{b}{{\sin {\rm{60}}^\circ }} = \frac{{10}}{{\sin {\rm{90}}^\circ }}\]$

Solving the equation $\[\frac{a}{{\sin 30^\circ }} = \frac{{10}}{{\sin {\rm{90}}^\circ }}\]$ to find the value of $a$:

$\[\begin{array}{l}\frac{a}{{\sin 30^\circ }} = \frac{{10}}{{\sin {\rm{90}}^\circ }}\\ \Rightarrow a = \frac{{10}}{{\sin {\rm{90}}^\circ }} \times \sin 30^\circ \\ \Rightarrow a = \frac{{10}}{1} \times \frac{1}{2}\\ \Rightarrow a = 5\end{array}\]$

Solving the equation $\[\frac{b}{{\sin {\rm{60}}^\circ }} = \frac{{10}}{{\sin {\rm{90}}^\circ }}\]$ to find the value of $b$:

$\[\begin{array}{l}\frac{b}{{\sin {\rm{60}}^\circ }} = \frac{{10}}{{\sin {\rm{90}}^\circ }}\\ \Rightarrow b = \frac{{10}}{{\sin {\rm{90}}^\circ }} \times \sin 60^\circ \\ \Rightarrow b = \frac{{10}}{1} \times \frac{{\sqrt 3 }}{2}\\ \Rightarrow b = 5\sqrt 3 \end{array}\]$

Hence, the value of $a$ is $5$ and the value of $b$ is $5\sqrt{3}$.

#SPJ2