Question

The question is on the image, please help quickly!

Answer 1

○
`c \approx \bf4748`

○
`\alpha \approx \bf 70.9^\circ`

○
`\beta = \bf 11.1^\circ`

Explanation:

Use the attached diagram for help.

• To solve for c, we have to use the cosine rule, where:

`c = √(a^2 + b^2 - 2ab \space\ cos \gamma)`

Substituting the values:

⇒
`c = √(4530^2 + 924^2 -2(4530)(924) * cos(98^\circ))`

⇒
`c \approx \bf4748`

• Now that we know the value of c, we can use the sine rule to calculate the value of α (alpha):

`(sin \space\ \alpha)/(a) = (sin \space\ \gamma)/(c)`

Substituting the values:

`(sin \space\ \alpha)/(4530) = (sin \space\ 98^\circ)/(4748)`

⇒
`sin \space\ \alpha = (sin \space\ 98^\circ)/(4748) * 4530`

⇒
`sin \space\ \alpha = 0.9448`

⇒
`\alpha \approx \bf 70.9^\circ`

• Since we now have the values of both ∠α and ∠γ, we can find the value of ß using the triangle sum theorem:

`\alpha + \beta + \gamma = 180^\circ`

⇒
`70.9^\circ + 98^\circ + \beta = 180^\circ`

⇒
`\beta + 168.9^\circ = 180^\circ`

⇒
`\beta = 180^\circ - 168.9^\circ`

⇒
`\beta = \bf 11.1^\circ`