Question

the adjoining figure, ABC is a triangle in
which angle B = 45° and angle C = 60°. If AD perpendicular on BC and
BC = 8 m, find the length of the altitude AD.​

Answer 1

Answer:

do it by taking tan 45 and tan 60 and then equate. you'll get -4(√3-3)m

Answer 2

Given:

ABC is a triangle

∠B = 45°

∠C = 60°

AD ⊥ BC

BC = 8cm

To Find:

the length of the altitude of AD.

Solution:

It is given that in ΔABC, ∠B = 45°

So,

In ΔABD,

tan45 = AD/BD [base/height]

     As we know, the value of tan45 is 1.

So, 1 = AD/BD

This implies, AD = BD..(i)

Then, ∠C = 60° it is given

So, In ΔADC,

tan60 = AD/DC [base/height]

We already know that the value of tan 60 is √3. So,

tan60 = AD/DC

√3   = AD/DC

DC = AD/√3..(ii)

Now, using the Pythagoras theorem,

⇒ a² + b² = c²

Now putting the values accordingly,

⇒ BD + DC = BC

It is given that BC = 8m

Then, substituting the values of DC and  BC.

⇒ AD + AD/√3 = 8   [ using(i) AD = BD and DC = AD/√3 using(ii)]

⇒ $\frac{\sqrt{3}AD + AD }{\sqrt{3} }$ = 8m

⇒ AD (√3+1) = 8√3

Now, solving the roots

⇒ AD = $\frac{8\sqrt{3} }{\sqrt{3}+1 }$

⇒ AD = $\frac{8\sqrt{3} }{\sqrt{3}+1 }*\frac{(1-\sqrt{3}) }{(1-\sqrt{3}) }$

⇒ AD = 24 - 8√3/3-1

Solving the denominator,

⇒ AD = 24 - 8√3/3-1

⇒ AD = 24-8√3/2

Then we will solve the numerator,

⇒ AD = 8(3 - √3)/2

⇒ AD = 4(3 - √3)

Therefore, the length of the altitude AD = 4(3 - √3).