Question

Given: △ABC, AB=5 sqrt2 m∠A=45°, m∠C=30° Find: BC and AC

Answer 1

Answer:

The measure of side BC = 5 unit and The measure of side AC = 5 unit

Step-by-step explanation:

Given as :

In a triangle ΔABC

The measure of side = AB = 5$\sqrt{2}$ unit

The measure of angle A = ∠A = 45°

The measure of angle C = ∠C = 30°

Now, For a triangle the sum of three angles of triangle = 180°

∴, ∠A + ∠B + ∠C = 180°

Or, 45° + ∠B +30° = 180°

Or , ∠B =  180° - ( 45° +  30°)

Or,  ∠B =  180° - 75°

i.e  ∠B =  105°

Now, From figure

Sin 45°  = $\dfrac{\textrm BC}{\textrm AB}$

Or, $\frac{1}{\sqrt{2} }$ = $\dfrac{\textrm BC}{5\sqrt{2}}$

Or, BC = $\frac{5\sqrt{2} }{\sqrt{2} }$

∴ BC = 5

So, The measure of side BC = 5 unit

Again , from figure

Cos 45°  = $\dfrac{\textrm AC}{\textrm AB}$

Or, $\frac{1}{\sqrt{2} }$ =$\dfrac{\textrm AC}{5\sqrt{2}}$

Or, AC = $\frac{5\sqrt{2} }{\sqrt{2} }$

∴  AC = 5

So, The measure of side AC = 5 unit

Hence The measure of side BC = 5 unit and The measure of side AC = 5 unit  Answer