Question

Please help i will give you 100 points

Answer 1

Answer:

x = 50.5°

Explanation:

In order to solve, get to know about the sine rule:

$\bf Sine \ Rule = \dfrac{A}{sinA} = \dfrac{B}{sinB}= \dfrac{C}{sinC}$

Solve for BD:

$\sf \dfrac{BD}{sin(50)} = \dfrac{5.6}{sin(78)}$

$\sf BD = \dfrac{5.6(sin(50))}{sin(78)}$

$\sf BD = 4.385686657 \ cm$

Then find angle D = 180° - 78° = 102°

Solve for angle C

$\sf \dfrac{4.385686657}{sinC} = \dfrac{9.3}{sin(102)}$

$\sf C = sin^{-1}(\dfrac{4.385686657sin(102)}{9.3} )$

$\sf C = 27.37 ^{\circ \:}$

Total Sum of interior angles of a triangle is 180°

$\sf B + C + D = 180^\circ$

$\sf x + 27.37^\circ + 102^\circ = 180^\circ$

$\sf x = 180^\circ - 102^\circ - 27.37^\circ$

$\sf x = 50.53^\circ$

$\sf x = 50.5^\circ \ \ \ (rounded \ to \ nearest \ 3 \ significant \ figure)$

Answer 2

Answer:

x = 50.5° (3 sf)

Step-by-step explanation:

Sine Rule for side lengths

$\sf \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

(where A, B and C are the angles and a, b and c are the sides opposite the angles)

Find BD:

$\implies \sf \dfrac{BD}{\sin BAD}=\dfrac{AB}{\sin BDA}$

$\implies \sf \dfrac{BD}{\sin 50^{\circ}}=\dfrac{5.6}{\sin 78^{\circ}}$

$\implies \sf BD=\dfrac{5.6\:sin 50^{\circ}}{\sin 78^{\circ}}$

$\implies \sf BD=4.385686657...cm$

Angles on a straight line sum to 180°

⇒ ∠ADB + ∠BDC = 180°

⇒ 78° + ∠BDC = 180°

⇒ ∠BDC = 102°

Sine Rule for angles

$\sf \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

(where A, B and C are the angles and a, b and c are the sides opposite the angles)

Find ∠BCD:

$\implies \sf \dfrac{\sin BCD}{BD}=\dfrac{\sin BDC}{BC}$

$\implies \sf \dfrac{\sin BCD}{4.385...}=\dfrac{\sin 102^{\circ}}{9.3}$

$\implies \sf BCD=\sin^{-1}\left(\dfrac{4.385...\sin 102^{\circ}}{9.3}\right)$

$\implies \sf BCD=27.46935172...^{\circ}$

The interior angles of a triangle sum to 180°

⇒ ∠CBD + ∠BDC + ∠BCD = 180°

⇒ x + 102° + 27.469...° = 180°

⇒ x = 50.53064828...°

⇒ x = 50.5° (3 sf)