Question

If an acute angle of a right triangle is 15° and hypotenuse is x cm then find the area of the right triangle.​

Answer 1

Your Answer:

Given:-

• angle B = 15°
• angle C = 90°
• AB = x units

To find:-

• Area of Triangle

Solution:-

$\tt \cos B = \dfrac{BC}{AB} \\\\ \tt \Rightarrow \cos {15}^{o}= \dfrac{BC}{x} \\\\ \tt \Rightarrow \dfrac{\sqrt{3}+1}{2\sqrt2} =\dfrac{BC}{x} \\\\ \tt \Rightarrow (\dfrac{\sqrt{3}+1}{2\sqrt2})x= BC$

and

$\tt \sin A = \dfrac{AC}{AB} \\\\ \tt \Rightarrow \sin {15}^{o}= \dfrac{AC}{x} \\\\ \tt \Rightarrow \dfrac{\sqrt{3}-1}{2\sqrt2} =\dfrac{AC}{x} \\\\ \tt \Rightarrow (\dfrac{\sqrt{3}-1}{2\sqrt2})x = AC$

Now area of triangle =

$\tt \dfrac{1}{2}\times base \times height \\\\ \tt = \dfrac{1}{2}\times BC \times AC \\\\ \tt = \dfrac{1}{2} \times (\dfrac{\sqrt3 + 1}{2\sqrt2})x \times (\dfrac{\sqrt3-1}{2\sqrt2})x \\\\ \tt =\dfrac{1}{2} \times \dfrac{(3-1)}{8} x^2 \rightarrow \rightarrow [\because (a+b)(a-b)=a^2-b^2]\\\\ \tt = \dfrac{1}{8}x^2$

So, area of triangle = (1/8)x^2

Answer 2

Question :–

If an acute angle of a right triangle is 15° and hypotenuse is x cm then find the area of the right triangle.

Given :–

• ∠B = 15°
• ∠C = 90°
• AB (Hypotenuse) = x

To Find :–

• Area of triangle.

Solution :–

First we need to find base (BC) & height (AC).

So,

1. Base (BC) :–

$→ \cos(b) = \frac{BC}{AB}$

★ b = 15° ★

★ AB = x ★

→ $cos \: 15 \degree = \frac{BC}{x}$

→ $\frac{ \sqrt{3} + 1 }{2 \sqrt{2} } = \frac{BC}{x}$

→ $( \frac{ \sqrt{3} + 1 }{2 \sqrt{2} } )x = BC$

→ $\boxed{ \sf \blue{ BC = \dfrac{\sqrt{3} + 1}{2 \sqrt{2}}}}$

2. Height (AC) :–

→ $\sin(a) = \frac{AC}{AB}$

★ a = 15° ★

★ AB = x ★

→ $\sin \: 15 \degree = \frac{AC}{x}$

→ $\frac{ \sqrt{3} - 1 }{2 \sqrt{2} } = \frac{AC}{x}$

→ $( \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } )x = AC$

→ $\boxed{ \sf \blue{ AB = \dfrac{\sqrt{3} - 1}{2 \sqrt{2}}}}$

Now,

Area of triangle → $\frac{1}{2} \times base \times height$

Now, put the value of base (BC) & height (AC) on the formula of triangle.

→ $\frac{1}{2} \times BC \times AC$

→ $\frac{1}{2} \times (\frac{ \sqrt{3} + 1 }{2 \sqrt{2} } )x \times (\frac{ \sqrt{3} - 1 }{2 \sqrt{2} } )x$

[(a + b)(a - b) = $\sf {a}^{2} - {b}^{2}$ ]

→ $\frac{1}{2} \times\frac{ (\sqrt{3} - 1 )}{8} {x}^{2}$

→ $\frac{1}{8} {x}^{2}$

Hence, the area of the triangle is $\boxed {\sf \red{\frac{1}{8} {x}^{2}}}$