Find out right the area???????

ABC is a isosceles triangle whose AB=AC=10cm. the angle BAC is 30 degrees. Find out the area of the triangle. ????

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Comments

Use the sine formula for the area of the triangle. $[ABC]=\frac{1}{2}(AB)(AC) \sin A$ .

Jaydee Lucero - 7 years, 8 months ago

by using this formula: Area=1/2(AB)(AC)sinA. area=1/2(10)(10)1/2=25 cm^2

Aya Anisa Dwinidasari - 7 years, 8 months ago

Or, if you don't want trigonometry, draw $AD$ perpendicular to $BC$ at $D$ . Since angles $B$ and $C$ both measure 75 degrees, you have no choice, but to use the ratio $\frac{AD}{AB}=\frac{AD}{AC}=\frac{\sqrt{2}+\sqrt{6}}{4}$ which still comes from trigonometry.

Jaydee Lucero - 7 years, 8 months ago

25cm^2 as the formula is (1/2)(10)(10)(sin 30)=25 (sin 30 =1/2)

Divyansh Singhal - 7 years, 8 months ago

Using cosine law you will determine BC.After that using hero's formula you will determine the area of triangle

EDSEL SALARIOSA - 7 years, 8 months ago

area=1/21010*sin30=25 cm^2

Nasir Mehmood Alam - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$