Question

Diagonals of a rhombus BEST intersect at A.
(i) f m∠B T S = 110 degree, then find m∠TBS
(ii) If l(TE) = 24, l(BS) = 70, then find l(TS) = ?

Answer 1

★ Given Information:

$\implies \sf{m \angle B T S = 100^\circ}$

$\implies l\sf{(TE) = 24}$

$\implies l \sf{(BS) = 70}$

★ To Find:

$\implies \sf{m \angle TBS}$

$\implies l \sf{(TS) }$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Let us draw the rough figure of ◻ BEST and show point A.

   

$\sf{(i) \; Opposite \; angles \; of \; a \; rhombus \; are \; congruent. }$

$\underline{\boxed{\sf{\therefore \color{pink}{\; m \angle BES = m \angle B T S = 110}}}}$

   

Now,

$\implies \sf{m \angle B T S + m \angle BES + m \angle TBE + m \angle TSE = 360^\circ}$

$\implies \sf{110^\circ + 110^\circ + m \angle TBE + m \angle TSE = 360^\circ}$

$\implies \sf{220^o + m \angle TBE + m \angle TSE = 360^\circ}$

$\implies \sf{m \angle TBE + m \angle TSE = 360^\circ - 220^\circ = 140^\circ}$

$\implies \sf{2m \angle TBE = 140^\circ } \dots \Big \langle {\frak{\blue{\because \; Opposite \; angles \; of \; a \; rhombus \; are \; congruent}}} \Big \rangle$

$\implies \sf{m \angle TBE = \dfrac{\cancle{140^\circ}}{\cancel{2}}}$

$\implies \sf{m \angle TBE = 70^\circ}$

$\implies \sf{m \angle TBE = \dfrac{1}{\cancel{2}} \times \cancel{70^\circ} = 35^\circ } \dots \Big \langle {\frak{\blue{\because \; diagonal \; of \; a \; rhombus\; bisects \; the \; opposite \; angles}}} \Big \rangle$

   

$\sf{(ii) \; Diagonals \; of \; a \; rhombus \; are \; perpendicular \; bisectors \; of \; each \; other.}$

∴ In ∆ TAS,

$\underline{\boxed{\sf{\color{pink}{ m \angle TAS = 90^\circ }}}}$

   

$\implies l \sf{(TA) = \dfrac{1}{2} } \; \textit{l} \sf{(TE) = \dfrac{1}{\cancel{2}} \times \cancel{24} = 12}$

$\implies l \sf{(AS) = \dfrac{1}{2}} \; \textit{l} \sf{(BS) = \dfrac{1}{\cancel{2}} \times \cancel{70} = 35 }$

   

By Pythagora's Theorem,

$\implies l \sf{(TS)^2 =} \; \textit{l} \sf{(TA)^2 } + \textit{l} \sf{(AS)^2 = (12)^2 + (35)^2 = 144 + 1225 = 1369}$

$\implies \; \therefore \; l\sf{(TS) = \sqrt{1369 \;} = 37 }$

     

Final Answer:

$\red{\longmapsto}$ m∠TBS = 35° & l(TS) = 37°

Answer 2

Answer:

⟼ m∠TBS = 35° & l(TS) = 37°

Step-by-step explanation:

we know that, the Opposite angles of a rhombus are congruent.

m∠BES = m∠ᏴᎢᏚ = 110

⟹ m∠ᏴᎢᏚ + m∠BES + m∠TBE + m∠TSE = 360°

⟹ 110° + 110° + m∠TBE + m∠TSE = 360°

⟹ 220° + m∠TBE + m∠TSE = 360°

⟹ m∠TBE + m∠TSE = 360° − 220° = 140°

⟹ 2m∠TBE = 140° …⟨ ∵ Opposite angles of a rhombus are congruent ⟩

⟹ m∠TBE = 70°

⟹ m∠TBE = ½ × 70° = 35°

...⟨ ∵ diagonal of a rhombus bisects the opposite angles ⟩

we know that, Diagonals of a rhombus are perpendicular bisectors of eachother.

∴ In ∆ TAS,

m∠TAS = 90°

⟹ l(TA) = ½ l(TE) = ½ × 24 = 12

⟹ l(AS) = ½ l(BS) = ½ × 70 = 35

By Pythagora's Theorem,

⟹ l(TS)² = l(TA)² + l(AS)² = (12)² + (35)² = 144 + 1225 = 1369

⟹ ∴ l(TS) = √1369 = 37

⟼ m∠TBS = 35° & l(TS) = 37°