Question

In ΔQRS, s = 2.3 inches, ∠S=51° and ∠Q=44°. Find the area of ΔQRS, to the nearest 10th of an square inch.

Answer 1

2.4

Explanation:

Answer 2

Area of ΔQRS = 2.3 square inches

Explanation:

From the given information,

<S + <Q + <R =
`180^(o)`

51 + 44 + <R =
`180^(o)`

95 + <R =
`180^(o)`

<R =
`180^(o)` - 95

=
`85^(o)`

<R =
`85^(o)`

Applying the Sine rule, we have;

`(q)/(SinQ)` =
`(r)/(SinR)` =
`(s)/(SinS)`

Using
`(r)/(SinR)` =
`(s)/(SinS)`

`(r)/(Sin 85)` =
`(2.3)/(Sin51)`

r =
`(2.3*Sin85)/(sin51)`

= 2.9483

r = 2.9 inches

Also,
`(q)/(SinQ)` =
`(s)/(SinS)`

`(q)/(Sin44)` =
`(2.3)/(Sin51)`

q =
`(2.3*Sin44)/(Sin51)`

= 2.0559

q = 2.0 inches

From Herons formula,

Area of a triangle =
`√(s(s-q)(s-r)(s-s))`

s =
`(2.3 + 2.0 + 2.9)/(2)`

= 3.6

Area of ΔQRS =
`√(3.6(3.6-2.0(3.6-2.9)(3.6-2.3))`

= 2.2895

Area of ΔQRS = 2.3 square inches