The length of one side of a triangle is $\left(\sqrt{3}+1\right)\text{ cm}.$ The two interior angles on that side are $30^\circ$ and $45^\circ.$

What is the area of the triangle?

The answer is 1.36602.

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Nice solution!

A Former Brilliant Member
- 5 years, 4 months ago

I did it same way........!

Siva prasad
- 5 years, 4 months ago

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Nice way Bhaiya but there may have been other possibilities like if the side $\sqrt{3}+1$ is opposite to $30^{o}$ , then the area may change?

Department 8
- 5 years, 4 months ago

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But it is given that the included side is $\sqrt 3$ +1 that means the side is between the two angles.. !

Rishabh Jain
- 5 years, 4 months ago

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Um... it says that a side and two interior angles?

Department 8
- 5 years, 4 months ago

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@Department 8 – The ques says...The two interior angles on that side ....

Rishabh Jain
- 5 years, 4 months ago

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@Rishabh Jain – Oh never saw that!

Department 8
- 5 years, 4 months ago

@Rishabh Jain – Try this

Department 8
- 5 years, 4 months ago

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*
B=45° and /
*
C=30°. Also BC=√3+1.
By sine rule,
BC sinC=AB sinA. [/_A=105°]
=> AB=√2.
Similarly we get,AC=2. [sin105°=(√3+1)/2√2]
Now,area=1/2×(sinA×AB×AC)
=(√3+1)/2
=1.366

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Let the triangle be $\triangle ABC$ where $\angle ABC=30^\circ$ and $\angle ACB=45^\circ$ .

Let $AD$ be the altitude drawn to $BC$ .

So,

$\begin{aligned} \tan 45^\circ & =1\Rightarrow AD=DC \\ \tan 30^\circ & = \frac{1}{\sqrt{3}} \Rightarrow \frac{AD}{\sqrt{3}+1-AD}=\frac{1}{\sqrt{3}} \\ \sqrt{3}+1-AD & =\sqrt{3} AD \\ AD=1 \\ \text{area}(\triangle ABC) & =\frac{1}{2}\cdot 1 \cdot (\sqrt{3}+1) \\ & = \boxed{1.6602} \end{aligned}$